Differentiation: past paper questions & worked solutions

\paragraph{Solution} Differentiate $x=e^{2\tan y}$ with respect to $y$:

This is positive on the stated interval, so

Taking logarithms of the original relation gives $\ln x=2\tan y$. Using $\sec^2y=1+\tan^2y$,

Hence $a=2$, $b=4$ and $n=2$.

Use $\sin^2(3x)=(1-\cos6x)/2$:

The derivative of $(x^2+4)^{5/2}$ is $5x(x^2+4)^{3/2}$. Therefore

For example, differentiating this answer gives

which recovers the integrand. Each indefinite integral has its own arbitrary constant.

Apply the product rule:

At a stationary point, $y'=0$. Since $e^{3x}\gt 0$ for every real x, $6x+17=0$, which determines the unique stationary x coordinate.

The denominator derivative is $6x-4=2(3x-2)$. Hence reverse the logarithmic chain rule:

The log argument is always positive because $3x^2-4x+5=3(x-2/3)^2+11/3$.

Put $u=e^{2x}-1$, so $du=2e^{2x}\,dx$. Then $\int e^{2x}(e^{2x}-1)^{-3}\,dx=(1/2)\int u^{-3}\,du=-1/(4u^2)+C$. The source excludes $x=0$, where the integrand is undefined.

Factor $x^2+x-12=(x+4)(x-3)$ and combine the fractions:

The cancellation is valid because $x\gt 3$. In particular, the numerator simplifies as $49x+7x^2-21x=7x^2+28x=7x(x+4)$.

Rewrite $f(x)=7+21/(x-3)$. Differentiating gives

Differentiate $x=2y^2+5y-6$ with respect to $y$:

A tangent parallel to the $y$-axis is vertical, so use $dx/dy=0$:

Substituting into the curve gives

The logarithm is base 10. Thus $10^y=2x+1$ and $2x=10^y-1$.

Differentiate the result of (a) with respect to y:

Take the reciprocal and use $10^y=2x+1$:

The original logarithm requires $x\gt -1/2$.

Differentiate x with respect to y:

Since $0\lt 2y\lt \pi/2$, both sine and cosine are positive. Using $\sin^22y=x/6$ and $\cos^22y=1-x/6$,

On $0\lt x\lt 6$ this is nonzero, so take its reciprocal.

At $y=\pi/4$,

Differentiate $x$ with respect to $y$ using the chain rule:

At $P$, $dx/dy=2(3/2)(-6/\pi)=-18/\pi$. Since this is nonzero, the tangent gradient is $\dfrac{dy}{dx}=-\dfrac\pi{18}$, and its equation is

Put $y=0$ for its intersection with the $x$-axis:

Use the product rule and $\frac{d}{dx}\ln(2x)=1/x$:

Write $f(x)=e^{2x}(4x+3)^{-2}$. Product and chain rules give

Factoring out $2e^{2x}(4x+3)^{-3}$,

Thus $a=4$ and $b=-1$.

Apply the quotient rule, retaining the restriction $x\ne-3$:

The numerator vanishes at $10-4x=0$, so $x=5/2$; the denominator changes sign at the excluded value $x=-3$. For $x\lt -3$, the numerator is positive and denominator negative, so $y'\lt 0$. For $-3\lt x\lt 5/2$, both are positive, so $y'\gt 0$. For $x\gt 5/2$, the numerator is negative and denominator positive, so $y'\lt 0$. Thus the positive-gradient interval is as follows.

At $t=0$, $\log_{10}N=2$, so $N=10^2$.

Exponentiating base 10 gives

Thus $a=100$ and $b=10^{0.35}=2.238721\ldots$, which is $2.24$ to 2 decimal places.

Retain the unrounded base $10^{0.35}$ when differentiating:

At $t=5$ this is $0.35\ln(10)\,10^{3.75}=4531.9356\ldots$ bacteria per hour.

With $y=(2+3\sin x)/(\cos x+\sin x)$, the quotient rule gives

Expand the numerator and cancel the opposite mixed terms:

The denominator is $1+2\sin x\cos x$. Divide numerator and denominator by $\cos x$ to obtain

This displayed form is used where its trigonometric terms are defined; the preceding quotient form also covers points with $\cos x=0$ when the original denominator is nonzero.

Differentiate the sixth power and multiply by the derivative of $3x-2$: $dy/dx=6(3x-2)^5\cdot3$.

At $P=(1/3,1)$, the tangent gradient is $18(-1)^5=-18$. The normal gradient is $-1/(-18)=1/18$. Thus

Rearranging with integer coefficients gives the required equation.

Since $d(3x^2+k)/dx=6x$, the numerator is $3/2$ times this derivative. The denominator is positive because $k\gt 0$, so

Apply the limits to (a) and combine the logarithms:

Hence $\ln((75+k)/(12+k))=(2/3)\ln8=\ln4$, so $75+k=4(12+k)$. This gives $27=3k$ and $k=9$, consistent with $k\gt 0$.

Raise the power by one, divide by the new power and by the inner derivative 2:

The derivative of $1+2\cos x$ is $-2\sin x$, so a primitive is $-2\ln(1+2\cos x)$. The log argument is positive throughout $0\le x\le\pi/3$. Hence

Write the integrand as $8(2x-3)^{-3}$. A primitive is $-2(2x-3)^{-2}$, because its derivative is $8(2x-3)^{-3}$. Thus

Since $d(x^2+3)/dx=2x$, reverse the chain rule:

Taking base-10 logarithms gives

The vertical intercept is $2.5$ and the gradient is $\dfrac{3.25-2.5}{10-0}=0.075$. Therefore

To 3 significant figures, $\boxed{a=316,\quad b=1.19}$.

Each year the model multiplies the number of recycling households by $b$. This means an increase of approximately $\boxed{19\%\text{ per year}}$ (about $18.85\%$ using the unrounded value).

Retain the unrounded constants:

At $t=5$ this is $129.5021421\ldots$, hence

to 2 significant figures.

Using the chain rule,

Treat $k$ as a constant when integrating:

The stated inequality therefore implies

Since the exponential is increasing and $k+3\gt 0$,

The roots are $(21\pm\sqrt{697})/2$, so the positive upper bound is $k\lt (21+\sqrt{697})/2=23.700378782\ldots$. With $k\gt 1$ an integer, the greatest possible value is $k=23$.

\paragraph{Solution} Put $u=x^2+(3k-2)x$, so $u'=2x+3k-2$. By the product and chain rules,

As $2e^u\gt 0$, stationary points satisfy

For two distinct real roots, the discriminant must be positive:

Consequently

For these values the quadratic changes sign at both simple roots, so both stationary points are genuine turning points. Equality gives a repeated root with no derivative sign change and is excluded.

At $t=0$, $18=10+Ae^0=10+A$.

At $t=45$, $16=10+8e^{-45B}$, so $e^{-45B}=6/8=3/4$. Taking logarithms,

Differentiate $T=10+8e^{-Bt}$:

At $t=2$, using the unrounded value $B=\ln(4/3)/45$, $dT/dt=-0.05049372834\ldots$. The negative sign means the temperature is falling.

For every finite $t\ge0$, $8e^{-Bt}\gt 0$ because $B\gt 0$. Thus $T=10+8e^{-Bt}\gt 10$; as $t\to\infty$, T approaches 10 from above. It therefore cannot reach 5 degrees Celsius.

Since $2e^{3\sin x}\gt 0$, zeros occur when $\cos x=0$. The labelled point R is the right-hand crossing, at $x=3\pi/2$.

For $f(x)=2e^{3\sin x}\cos x$, the product and chain rules give

At a turning point the derivative is zero. The exponential factor is nonzero, so $3\cos^2x-\sin x=0$. Substitute $\cos^2x=1-\sin^2x$ and rearrange: $3-3\sin^2x-\sin x=0$.

Solving $3s^2+s-3=0$ for $s=\sin x$ gives

The negative root is less than $-1$, so it is impossible for a real sine. Let $\beta=\arcsin((\sqrt{37}-1)/6)$ in the first quadrant. The labelled minimum Q lies in the second quadrant, hence $x_Q=\pi-\beta=2.131037293\ldots$.

Initially $t=0$ and $N=32$, because $N$ is measured in thousands.

For 100000 phones, use $N=100$:

Taking natural logarithms,

Differentiate the model:

At $t=7$, this is $4.7265337\ldots$ thousand phones per month. Multiplying by 1000 gives $4726.5337\ldots$ phones per month.

For $t\ge0$, $93e^{-0.109t}\gt 0$, so $N\lt 125$ and $N\to125$ as $t\to\infty$. The model can therefore never reach 150000 phones: its limiting count is 125000.

Set $u=x^2+k\gt 0$. Applying the quotient and chain rules,

Comparing with the required form identifies both constants.

The denominator is positive, so stationary points satisfy

There are three distinct real stationary points precisely when $e-k\gt 0$. In that case the factor $1-\ln(x^2+k)$ is positive between the outer roots and negative outside them; $2x$ changes sign at zero. Thus the derivative changes sign at all three points, so all are turning points. Together with $k\gt 0$, the three coordinates below are real and distinct; the restriction is $0\lt k\lt e$.

At $k=e$, the two outer stationary points merge with $x=0$, so there are no longer three distinct turning points. Values of $k$ can approach $e$ from below, but cannot equal it under this condition.

Differentiate the logarithm directly on its domain $\sin3x\ne0$:

This does not assume $\sin3x\gt 0$; the logarithm's argument is its square.

Apply the power and chain rules:

From part (ii)(a), an antiderivative of $x(3x^2-4)^5$ is $(3x^2-4)^6/36$.

By the product rule,

Since the exponential is positive, stationary points have $x=2$ or $x=4/3$. The sign of $f'$ changes from positive to negative at $4/3$, so this is the maximum A.

The curve is nonnegative and has just one zero, $x=2$; hence $k=0$ is excluded. It tends to 0 from above as $x\to-\infty$, rises to $4e^4/9$, falls to 0, then rises without bound. For $0\lt k\lt 4e^4/9$ there are three intersections. At the maximum level there are two distinct intersections, including A. For a larger positive k there is only one.

Differentiate x with respect to y by the product rule:

Where $dx/dy\ne0$, take the reciprocal:

For $y\ne0$, the original relation gives $e^{2y}=x/y$, hence the required form $dy/dx=y/[x(1+2y)]$. At the origin, that last quotient is $0/0$; the preceding uncancelled derivative gives $dy/dx=1$. At $y=-1/2$ there is no finite $dy/dx$.

Consider $x=h(y)=ye^{2y}$. Its derivative $h'(y)=e^{2y}(1+2y)$ vanishes at $y=-1/2$. It is negative before this y value and positive after it, so the minimum x value is $h(-1/2)=-1/(2e)$. As $y\to-\infty$, $h(y)\to0$ from below; also $h(0)=0$ and h then increases without bound. Thus a level $x=k$ meets both negative-y branches precisely between the minimum and zero. At the minimum the two intersections coincide; at zero there is just the finite point $y=0$.

The principal arcsine curve increases from $(-2,-\pi/2)$ through $(0,0)$ to $(2,\pi/2)$. It is concave down on the negative-x half, concave up on the positive-x half, and vertical at both ends.

Differentiate the supplied inverse relation: $dx/dy=2\cos y$. For $-\pi/2\lt y\lt \pi/2$, cosine is positive, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2/4}$. Taking the reciprocal gives

Thus $A=4$; the endpoints have vertical tangents, not finite derivatives.

At $y=\pi/4$, $x=2\sin(\pi/4)=\sqrt2$. The gradient is $1/\sqrt{4-2}=1/\sqrt2$. Hence $y-\pi/4=(x-\sqrt2)/\sqrt2$, so $y=x/\sqrt2+\pi/4-1$.

Using the product and chain rules,

At a stationary point, divide by $x/2$, since $x\gt 0$, to obtain $4\cos(x/2)=x\sin(x/2)$. The endpoint $x=\pi$ does not satisfy this equation, so at P we have $0\lt x\lt \pi$ and $\cos(x/2)\gt 0$. Hence $\tan(x/2)=4/x$. The principal arctangent is valid because $0\lt x/2\lt \pi/2$:

Use radian mode and retain full calculator precision between iterations:

Thus the requested second and sixth terms, not the limiting value, are as below.

Taking base-ten logarithms gives $\log_{10}A=\log_{10}p+t\log_{10}q$. The intercept is 0.32 and the gradient is $(0.56-0.32)/8=0.03$. Thus $p=10^{0.32}=2.08929613\ldots$ and $q=10^{0.03}=1.071519305\ldots$.

Using the parameter values requested in (a), $A=2.089(1.072)^t$. For an arbitrary positive base, $d(q^t)/dt=q^t\ln q$, so

This is a positive area increase per day. Retaining unrounded parameters also gives 0.22 to two decimal places.

Write $f(x)=(2x+3)(4x-1)^{-1/2}$. The product and chain rules give

The denominator of $f'(x)$ is positive throughout the domain, so the derivative is negative for $1/4\lt x\lt 2$ and positive for $x\gt 2$. Consequently $x=2$ is the global minimum, with $f(2)=7/\sqrt7=\sqrt7$. Also $f(x)\to\infty$ as $x\to1/4^+$ and as $x\to\infty$. By continuity all values at least $\sqrt7$ are attained, including the minimum.

Apply the product rule to the exponential term:

The derivative is continuous on $[0,1]$, and

The change of sign therefore places a zero of $f'$ between 0 and 1. Since the zero is unique, $\boxed{\alpha\in(0,1)}$, and hence it is in $[0,1]$.

Start at $x_1=0$, retaining unrounded values in each substitution:

Thus $\boxed{x_3=0.3260}$ to 4 decimal places.

Continuing the supplied iteration gives

Further iteration stabilises at $0.367471265\ldots$, so $\boxed{\alpha=0.3675}$ to 4 decimal places.

Using the quotient rule,

Thus $a=4$, $b=-24$ and $c=27$.

A stationary point requires

The roots are $3/2$ and $9/2$. Reject $9/2$ because the stated domain is $x\lt 3$. At $x=3/2$, $f(x)=-2$, so the turning point is $\boxed{(3/2,-2)}$. The derivative changes from positive to negative there, confirming a maximum.

The inside factor 2 halves the horizontal coordinate; adding 3 raises the vertical coordinate by 3. Therefore

At $t=0$, $N=4000/(19+1)=200$ squirrels.

The quotient rule gives

At $t=T$, the derivative is zero. The exponential and denominator are positive, so $19-e^{0.2T}=0$. Therefore $e^{0.2T}=19$ and $A=19$. The derivative is positive before $T$ and negative after $T$, confirming a maximum.

Since $e^{0.2T}=19$, we have $e^{0.1T}=\sqrt{19}$. Substitution into the model gives

The maximum number is $459$ squirrels to the nearest whole number.

At $t=0$, $75=21+A$, so $\boxed{A=54}$.

At $t=5$, $25=21+54e^{-5k}$, hence

Differentiate the temperature model:

A decrease of $9\,{}^{\circ}\mathrm{C}$ per minute means $d\theta/dt=-9$, not $+9$. Thus

Using the unrounded value of $k$,

Since $x\gt 0$, $6\sqrt{x}$ is positive. Therefore $f(x)=0$ requires $\ln(4x)=0$, so $4x=1$ and $x=1/4$.

Apply the product and chain rules:

At $Q$, set $f'(x)=0$. Since $3/\sqrt{x}\gt 0$,

Then $\sqrt{x}=1/(2e)$ and $\ln(4x)=-2$, giving $f(x)=6(1/(2e))(-2)=-6/e$.

The factor $3/\sqrt{x}$ is positive and $\ln(4x)+2$ increases through zero at $Q$, so $f'$ changes from negative to positive: $Q$ is the global minimum. As $x\to\infty$, $f(x)\to\infty$, hence the range of $f$ is $[-6/e,\infty)$. Multiplication by $-2$ reverses the inequality and includes the maximum $12/e$.

At $t=0$, $2500=Ae^0$, so $A=2500$ exactly. At $t=8$, $10000=2500e^{8k}$; hence $e^{8k}=4$ and $k=\ln4/8=0.1732867951\ldots$. To four significant figures:

For the second population,

The requested rate of decrease is the positive magnitude, to three significant figures.

Retain the exact value $k=\ln4/8$. Equating populations gives

Take natural logs and solve:

The quotient rule gives, for $y\ne1$,

A tangent parallel to the $y$-axis is vertical, so $dx/dy=0$. The numerator gives $(y-3)(y+1)=0$, hence $y=3$ or $y=-1$; neither makes the denominator zero. Then

Thus $Q=(4,3)$ and $P=(-4/3,-1)$, with the vertical tangent equations below.

Exponentiate base 10:

Thus $p=10^{2.74}=549.5408739\ldots$ and $q=10^{0.079}=1.199499303\ldots$. Round p to the nearest integer and q to two significant figures.

At $t=0$, $x=pq^0=p$. Thus p is the initial amount of antibiotic in the patient's bloodstream, approximately 550 milligrams immediately after the dose in this model.

For the second patient, write $x=400e^{-t\ln1.4}$. Differentiate to obtain $dx/dt=-400\ln(1.4)\,1.4^{-t}$. At $t=5$, the derivative is $-25.02470969\ldots\,\mathrm{mg\,h^{-1}}$. The negative sign indicates the amount is decreasing.

Differentiate to obtain $y'=6x/(x^2-5)-8x$. At a stationary point in $x\gt \sqrt5$, x is nonzero, so $6/(x^2-5)=8$, giving $6=8x^2-40$ and $x^2=23/4$. The original positive domain selects $x=\sqrt{23}/2$, which exceeds $\sqrt5$. Hence $p=23$.

(a) Differentiate the squared sine with the chain rule: $y'=4-24\sin x\cos x=4-12\sin2x$. Thus $A=4$, $B=-12$. (b) Since $-1\le\sin2x\le1$, the largest derivative is attained when $\sin2x=-1$: $4-12(-1)=16$.

By the product and chain rules, $f'(x)=\cos(x/3)-(x/3)\sin(x/3)$.

Set $f'(x)=0$: $\cos(x/3)=(x/3)\sin(x/3)$. At a solution, cosine cannot vanish (that would leave a nonzero right side), so $\tan(x/3)=3/x$. On the principal positive branch $0\lt x/3\lt \pi/2$, this gives

This is the branch used by the supplied iteration for the root near 2.581. The original domain $x\gt 0$ also admits later tangent branches: in general $x=3\arctan(3/x)+3n\pi$. The principal-branch rearrangement must not be interpreted as listing every positive stationary point.

Using radians and $x_1=2.5$, the successive values are

Keep full precision between steps and round the requested terms only.

Define $F(x)=\cos(x/3)-(x/3)\sin(x/3)$, which is continuous near 2.581. The rounding interval endpoints give

By continuity a root lies strictly between these endpoints; every number in this open interval rounds to 2.581 to three decimal places.

At $t=5$, the growth rate is $180e=489.290729\ldots$ insects per year.

At $t=0$, $10000=8000+P$, so $P=2000$ exactly. Using the population at $t=4$,

Taking natural logarithms,

At equality, $8000+900e^{0.2T}=8000+2000e^{kT}$, hence

Use the unrounded value $k=\ln(1.785)/4$:

The gradient is $(2.25-2)/(5-0)=0.05$ and the vertical intercept is $2$. Hence $\log_{10}V=0.05t+2$.

Thus $a=100$ exactly and $b=10^{0.05}=1.122018454\ldots=1.12$ to three significant figures. In the requested rounded-parameter form, $V=100(1.12)^t$; retain the unrounded base for later calculations.

Differentiating the unrounded model,

At $t=T$ this equals $50$, so

Therefore $T=13$ years to the nearest integer.

Apply the product and chain rules:

Since $e^{-x^2}\gt 0$, a stationary point satisfies

Neither endpoint $x=0,\pi/3$ is stationary. In the interior $\sin3x\gt 0$ and $x\gt 0$, so $\cos3x\gt 0$ at a stationary point and $0\lt 3x\lt \pi/2$. Therefore division and the principal inverse tangent are valid:

Work in radians and retain calculator precision between iterations:

Thus (i) $\boxed{x_2=0.4367}$ and (ii) $\boxed{x_4=0.4307}$.

Let $h(x)=3\cos3x-2x\sin3x$, which is continuous. At the rounding boundaries,

The sign change places a root strictly between $0.4295$ and $0.4305$. On $0\lt x\lt \pi/6$, $h'(x)=-11\sin3x-6x\cos3x\lt 0$, so the stationary root is unique; for $\pi/6\leq x\leq\pi/3$, $h(x)\lt 0$. Hence the $x$ coordinate of $P$ is $\boxed{0.430\text{ to 3 decimal places}}$.

At $t=0$, $e^{-0.05t}=1$, so $H=280+24=304$ degrees Celsius.

The curve starts at $(0,304)$ and decreases for $t\ge0$. The exponential term stays positive and tends to zero, so the curve approaches $H=24$ from above without crossing it.

Substitute $H=144$:

Take natural logarithms: $-0.05t=\ln(3/7)$. Hence $t=-20\ln(3/7)=20\ln(7/3)=16.9459572077\ldots$ minutes.

Differentiate using the chain rule:

The original equation gives $280e^{-0.05t}=H-24$, so

Differentiate $x=3\cos2y$ with respect to $y$ using the chain rule:

For $0\lt y\lt \pi/2$, $\sin2y\gt 0$ and

Since $\cos2y=x/3$ and the sine is positive on this interval,

Consequently $dy/dx=-1/(2\sqrt{9-x^2})$, valid for $-3\lt x\lt 3$. The endpoint tangents are vertical; there is no finite derivative there.

Set the gradient equal to $-1/4$:

Because $a\gt 0$, take $a=\sqrt5$. Substitute into $a=3\cos2b$:

Here $0\lt 2b\lt \pi/2$ since both coordinates are positive, so the principal arccos value is the required one.

For $y=(4x-7)^{1/2}$,

At $P=(8,5)$, the tangent gradient is $2/5$, so the normal gradient is $-5/2$. The point-gradient equation is

The curve meets the $x$-axis at $x=7/4$. The normal meets it at $x=10$. Split the shaded region at $P$, into a curve area from $7/4$ to 8 and a triangle from 8 to 10:

The triangle has area $\frac12(10-8)\cdot5=5$. Hence the total area is $125/6+5$.

At $x=1/4$, $\sin^2 4y=1/4$. Since $0\le4y\le\pi/2$,

Only this branch lies in the given range of $y$.

Differentiate $x=(\sin4y)^2$ with respect to $y$:

On the interior of the given interval, $\sin4y=\sqrt{x}$ and $\cos4y=\sqrt{1-x}$ are positive. Taking the reciprocal,

Complete the square:

Thus $dy/dx=(q+r(x+s)^2)^{-1/2}$ has the constants below, with $0\lt x\lt 1$ for a finite derivative; the endpoints have vertical tangents.

The derivative is positive. Its minimum occurs when the square root in the denominator is largest. Since $(x-1/2)^2\ge0$, the largest radicand is 16, attained at $x=1/2$. Hence the smallest gradient is $1/\sqrt{16}$.

The product rule gives

Comparison with the stated form gives $A=2$.

Since $e^{-4x}\gt 0$, $f'(x)=0$ means $2x+1=0$ or $1-4x=0$. Thus $x=-1/2$ or $x=1/4$. Substitution gives $f(-1/2)=0$ and $f(1/4)=(3/2)^3e^{-1}=27/(8e)$.

The derivative changes from positive to negative at $x=1/4$, so this is the maximum; at $x=-1/2$ it does not change sign. The transformation $g(x)=8f(x-2)$ sends $(x,y)$ to $(x+2,8y)$. Hence the maximum stationary point is $(1/4+2,8\cdot27/(8e))=(9/4,27/e)$.

Put $u=y-\pi/6$. Since $x=3\tan u$,

This is positive for all real $x$, so we may take the reciprocal: $dy/dx=3/(x^2+9)$. Thus the required integers are $a=3$, $b=9$.

At $y=\pi/3$, $x=3\tan(\pi/3-\pi/6)=3\tan(\pi/6)=\sqrt3$. The tangent gradient is $3/((\sqrt3)^2+9)=1/4$. Its point-gradient equation is

At its x-axis crossing $y=0$, $-4\pi/3=x-\sqrt3$.

At the start $t=0$, so $e^{kt}=1$ and the model gives $F(0)=350/(9+1)=35$ fruit flies.

Using $F(15)=200$,

Take natural logs to obtain $15k=\ln12$, as required.

By the quotient rule,

Set $F'(T)=10$ and let $u=e^{kT}\gt 0$. Then

With the exact $k=\ln12/15$, the quadratic formula gives

Both values are positive and at least 1, hence both correspond to $T\ge0$. Finally $T=\ln u/k$ gives $5.6779528361\ldots$ or $20.8488936858\ldots$ days. To one decimal place the two times are as follows.

Differentiate by the quotient rule:

Inside the stated interval, $1+\sin x\gt 0$. At the interior minimum M, $y'=0$, so its numerator is zero, giving the required equation.

Write $2\sin x+\cos x=\sqrt5\sin(x+\beta)$, where $\cos\beta=2/\sqrt5$, $\sin\beta=1/\sqrt5$ and $\beta=\arctan(1/2)$. The equation becomes $\sin(x+\beta)=-2/\sqrt5$. Let $\gamma=\arcsin(2/\sqrt5)$, so $\gamma+\beta=\pi/2$. The solutions in the closed boundary range include $x=-\pi/2$, $3\pi/2$, and

Both boundary values are outside the original open domain (and make its denominator zero). The sole interior stationary point is M; the derivative changes from negative to positive there.

Let $F(x)=6\ln(2x+3)-x^2/2+4$. Then $F(-1.25)=-0.94013308\ldots$ and $F(-1.2)=0.21504626\ldots$. The log argument is positive on this interval, so F is continuous there. The change of sign proves a zero between these endpoints, giving the stated bracket for P.

Apply the given recurrence with $x_1=6$:

Continuing without rounding the internal values,

Further iteration settles at $6.41417418\ldots$. The positive square-root branch corresponds to Q.

Differentiate: $F'(x)=12/(2x+3)-x$. For a stationary point, $12=x(2x+3)$, or $2x^2+3x-12=0$. The quadratic formula gives $x=(-3\pm\sqrt{105})/4$. The negative choice is less than $-3/2$ and outside the original domain. Also $F”(x)=-24/(2x+3)^2-1\lt 0$, confirming the admissible point is a maximum.

Factor $x^2+7x+12=(x+3)(x+4)$. A common denominator gives

The cancelled factor is nonzero throughout the original domain $x\gt 0$.

Let $y=5x/(x+3)$. Then $yx+3y=5x$, so $x=3y/(5-y)$. Exchange variable names. Since $f(x)=5-15/(x+3)$ for $x\gt 0$, its range is $0\lt f(x)\lt 5$. This becomes the inverse domain.

(i) By the quotient rule, $f'(x)=[5(x+3)-5x]/(x+3)^2=15/(x+3)^2$. (ii) The numerator and denominator are positive on $x\gt 0$, so $f'(x)\gt 0$ throughout its domain. Therefore f is increasing.

Write $f(x)=(x^3-4x)e^{-x/2}$. By the product rule,

At O, $f'(0)=-4$, so the normal gradient is $1/4$ and its equation is $y=x/4$. At the other intersection P, x is nonzero; equating curve and normal and dividing by x gives $(x^2-4)e^{-x/2}=1/4$. Thus $x^2=4+e^{x/2}/4=(16+e^{x/2})/4$. Since P lies in the third quadrant, select the negative square root.

Starting from $x_1=-2$, $x_2=-\tfrac12\sqrt{16+e^{-1}}=-2.0228617996\ldots$. Further unrounded iterates are $x_3=-2.0226034084\ldots$, $x_4=-2.0226063126\ldots$ and $x_5=-2.0226062799\ldots$. The continuing sequence settles at $-2.0226062803\ldots$.

By the product and chain rules,

For $-\pi/4\lt x\lt \pi/4$, $e^{3x}\sec2x$ is positive and nonzero. Hence $g'(x)=0$ precisely when $3+2\tan2x=0$, or $\tan2x=-3/2$. Since $-\pi/2\lt 2x\lt \pi/2$, tangent is one-to-one there:

Differentiate $x=\ln(\sin y)$ with respect to y:

The original equation gives $\sin y=e^x$. Since $0\lt y\lt \pi/2$, cosine is positive, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-e^{2x}}$. Therefore

Here $x\lt 0$ from the original interval, so the square root is real and nonzero.

For $x\gt -7/4$, apply the product and chain rules:

The interior stationary values satisfy $x^2(2x+3)=0$, so $x=0$ or $x=-3/2$. Near $x=-3/2$, the derivative changes from negative to positive; near $x=0$ it stays positive on both sides. Thus the minimum turning point is at $x=-3/2$.

The endpoint has $f(-7/4)=0$, the global minimum is $-27/8$ and $f(x)$ tends to infinity as $x$ increases. Hence the range of $f$ is $[-27/8,\infty)$. Multiplication by $-4$ reverses the inequality and makes the upper endpoint $27/2$.

The transformation is a shift right by $3/2$, vertical stretch by 40, then shift down by 8:

Therefore $Q=(1/2,3/8)$ maps to $(1/2+3/2,\;40\cdot3/8-8)$.

To find $f^{-1}(22)$, solve $f(u)=22$: $2+5\ln u=22$, so $\ln u=4$ and $u=e^4\gt 0$.

Using the quotient rule,

For every $x\gt 1/3$, the denominator is positive, so $g'(x)\gt 0$. Thus g is strictly increasing on its entire domain.

Set $y=(6x-2)/(2x+1)$. Then

As $x\downarrow1/3$, $g(x)\to0$; as $x\to\infty$, $g(x)\to3$. Since g is strictly increasing, its range is $(0,3)$; neither limiting value is attained. Interchange the input and output symbols.

The inner function g takes every value in $(0,3)$. Since $f(u)=2+5\ln u$ is increasing for $u\gt 0$, its values on this interval run from $-\infty$ to, but not including, $2+5\ln3$.

At the first landing, $h=0$ and $x=d\gt 0$:

Taking logarithms, $0.02d=\ln3$, hence $d=50\ln3=54.930614\ldots$. The root $x=0$ is the launch point, not the landing point.

By the product and chain rules,

At a stationary point,

Since $x+50\gt 0$, logarithms give $0.02x=\ln(150/(x+50))$. Also $h”(x)=-(0.02+0.0002x)e^{0.02x}\lt 0$ on $[0,d]$. Thus the stationary point is the unique maximum; the endpoints both have zero height.

(i) With $x_1=30$,

(ii) Continuing with full intermediate precision gives $x_3=30.54431108\ldots$, $x_4=31.09139065\ldots$, $x_5=30.75292479\ldots$ and $x_6=30.96205562\ldots$. Continued iteration converges to $30.88212\ldots$, which lies in $(0,d)$. This is the horizontal distance at maximum height, not the maximum height itself.

At $y=\pi/3$, $x=4(\sqrt3/2)^2-1=3-1=2$.

(i) Differentiating with respect to $y$ gives

(ii) For $0\lt y\lt \pi/2$, both sine and cosine are positive, and $-1\lt x\lt 3$. From $x+1=4\sin^2y$,

Consequently $dx/dy=2\sqrt{x+1}\sqrt{3-x}$. Taking the reciprocal proves the required formula on $-1\lt x\lt 3$. At $x=-1$ and $x=3$ the denominator is zero: these are endpoint vertical tangents, not finite derivatives.

At $P=(2,\pi/3)$ the tangent gradient is $1/(2\sqrt3)$, so the normal gradient is $-2\sqrt3$. Thus the normal is $y-\pi/3=-2\sqrt3(x-2)$. Setting $y=0$ gives $N=(2+\pi/(6\sqrt3),0)$. Triangle OPN has horizontal base $ON=2+\pi/(6\sqrt3)$ and perpendicular height $\pi/3$:

The exponential is positive and the squared factor is nonnegative, so $f(x)\ge0$. The value zero is attained at $x=\pm\sqrt{3/2}$. The graph has its highest point at $x=0$, where $f(0)=9$; the outer maxima are lower. Thus both endpoints of the range are included.

For $f(x)=e^{-x^2}(2x^2-3)^2$, product and chain rules give

Factoring yields

The stationary abscissae are $0$, $\pm\sqrt{3/2}$ and $\pm\sqrt{7/2}$. At the outer local maxima,

The central maximum is 9 and the two minima are zero. Also $f(x)\to0$ as $x\to\pm\infty$. A positive horizontal line above both outer maxima but below 9 meets only the two sides of the central peak. At $k=16e^{-7/2}$ it also touches both outer peaks, giving four distinct intersections; at $k=9$ there is only one. Both bounds must be excluded.

Use product and chain rules:

With a common denominator,

This derivative formula applies for $x\gt -9/4$; the endpoint derivative is not finite.

The denominator is positive in the derivative domain. Thus $5x^2+9x-2=(5x-1)(x+2)=0$.

The derivative changes from positive to negative at $x=-2$, so this is the local maximum P. Substitute into f: $f(-2)=5(4-2)\sqrt{-8+9}=10$.

On $[-9/4,0]$, f rises from $f(-9/4)=0$ to $f(-2)=10$, then decreases to $f(0)=-30$. The other stationary point $1/5$ is outside this restricted interval. Thus the minimum and maximum of f on it are $-30$ and 10. Applying the positive vertical transformation $g=2f+4$ gives endpoints $2(-30)+4=-56$ and $2(10)+4=24$, both attained.

Apply the chain rule twice:

From $x=3\sec^22y$, $\sec^22y=x/3$ and $\tan^22y=x/3-1$. Since $0\lt 2y\lt \pi/2$, tangent is positive, so $\tan2y=\sqrt{(x-3)/3}$. Thus $dx/dy=4x\sqrt{(x-3)/3}\gt 0$. Taking the reciprocal gives

At $y=\pi/12$, $2y=\pi/6$ and $x=3(4/3)=4$. The tangent gradient is $\sqrt3/(4\cdot4\sqrt1)=\sqrt3/16$. Therefore the normal gradient is $-16/\sqrt3=-16\sqrt3/3$. Using the point $(4,\pi/12)$,

On the $y$ axis, $x=0$, so $\sin(3y+\pi/4)=0$. Since $\pi/12\lt y\lt 3\pi/4$, we have $\pi/2\lt 3y+\pi/4\lt 5\pi/2$. Therefore the possible angles are $\pi$ and $2\pi$, giving

Differentiate $x$ with respect to $y$, then take the reciprocal:

Using $\sin(3y+\pi/4)=3x/2$,

Thus $\boxed{p=4,\ q=9}$. The derivative formula applies where the tangent is not vertical.

At $x=0$, part (b) gives $|dy/dx|=1/2$. The sketch shows a negative tangent gradient at $A$ and a positive one at $B$. The normal at $A$ therefore has gradient $2$, while the tangent at $B$ has gradient $1/2$:

At their intersection,

Differentiate $y=(3+4e^{x^2})^{1/2}$ using the chain rule:

Since the tangent at $P=(\alpha,\sqrt{3+4e^{\alpha^2}})$ passes through the origin, its gradient is $y_P/\alpha$, with $\alpha\gt 0$. Equate this to the derivative:

Consequently $x=\alpha$ solves the displayed equation.

Let $h(x)=4x^2e^{x^2}-4e^{x^2}-3$, a continuous function. Then $h(1)=-3\lt 0$ and $h(2)=12e^4-3=652.1778\ldots\gt 0$, so a root lies between 1 and 2. Moreover $h'(x)=8x^3e^{x^2}\gt 0$ for $x\gt 0$, so this is the unique positive root corresponding to the tangent abscissa $\alpha$.

From $4x^2e^{x^2}=4e^{x^2}+3$, divide by the positive $e^{x^2}$ to get $4x^2=4+3e^{-x^2}$. The source domain has $x\ge0$, so take the positive square root.

With $x_1=1$, the unrounded iterates start

Continue: $x_6=1.1051212304\ldots$, $x_7=1.1050504628\ldots$, $x_8=1.1050661137\ldots$, $x_9=1.1050626523\ldots$; further iteration settles at $1.1050632792\ldots$. Therefore (i) $x_3=1.0997$ and (ii) $\alpha=1.1051$, both to four decimal places.

By the quotient rule,

For every $x\gt 4$, the denominator is positive and the numerator is negative. Therefore $f'(x)\lt 0$ throughout its domain, so f is decreasing.

Set $y=(5x-3)/(x-4)$. Then $yx-4y=5x-3$, giving $x(y-5)=4y-3$ and $x=(4y-3)/(y-5)$. Exchange the variable names. Since $f(x)=5+17/(x-4)$ on $x\gt 4$, its range is $(5,\infty)$, which becomes the inverse's domain.

(i) For $x\gt 4$, $f(x)\gt 5\gt 4$, so the second application is defined.

Thus $a=22$, $b=-3$, $c=13$. (ii) Rewrite $ff(x)=22-289/(x+13)$ with the original domain $x\gt 4$. It is strictly increasing, tends to 5 as $x\to4^+$ and to 22 as $x\to\infty$. Neither limiting value is attained.

At $t=0$, $18=A-180e^0=A-180$. Hence $A=198$.

At $t=5$, $90=198-180e^{-5k}$, so $180e^{-5k}=108$ and $e^{-5k}=3/5$. Taking logarithms gives $-5k=\ln(3/5)$, hence $k=(1/5)\ln(5/3)$. Thus one suitable rational pair is $p=1/5$, $q=5/3$.

Use the unrounded k:

Round the temperature to three significant figures.

Differentiate with respect to time: $d\theta/dt=180ke^{-kt}$. At $t=9$ this is $180[\ln(5/3)/5]e^{-(9/5)\ln(5/3)}=7.332421555\ldots$. The sign is positive because the oven is warming.

Using the quotient rule and the chain rule,

Thus $P(x)=(2x+5)(2x-17)$ and $Q(x)=(x-3)^2$ are fully factorised quadratics.

For $x\ne3$, the denominator $(x-3)^2$ is positive, so the sign of $f'(x)$ is the sign of $(2x+5)(2x-17)$. For an increasing function we require

Both factors are negative when $x\lt -\frac52$, and both are positive when $x\gt \frac{17}{2}$. Therefore the required values are

For $0\lt x\lt \frac{\pi}{4}$, we have $\sin4x\gt 0$. Applying the product rule and the chain rule gives

Since $g(0)=0$ while $g(x)\gt 0$ in the interior, the stated maximum $M$ does not occur at $x=0$. Its coordinate therefore lies in $0\lt x\lt \frac{\pi}{4}$, where $g$ is differentiable, so $g'(x)=0$ at $M$. Consequently,

At such a point $\cos4x\ne0$: the only interior zero of $\cos4x$ is $x=\frac{\pi}{8}$, where the left-hand side above is $1$, not $0$. Dividing by $\cos4x$ therefore gives

Let $f(x)=2\cos3x-3x+4$, using radians. Then

The function is continuous on $[0.8,0.9]$, so this change of sign shows that there is a root between the endpoints. Moreover, $f'(x)=-6\sin3x-3\lt 0$ on this interval, so this root is unique there. Thus the crossing shown at $P$ has $\boxed{0.8\lt \alpha\lt 0.9}$.

Using the inverse cosine in radians,

Hence $\boxed{x_2=0.8327}$ to 4 decimal places.

Continue the iteration, retaining the full calculator values at each step:

Therefore $\boxed{x_5=0.8110}$ to 4 decimal places.

Differentiate the curve equation:

At a stationary point,

For $x\gt 0$, the stationary points therefore belong to the two families

In the first family, $\cos3x=-\frac{\sqrt3}{2}$, giving $y”=9\sqrt3\gt 0$, so these are local minima. In the second family, $\cos3x=\frac{\sqrt3}{2}$ and $y”=-9\sqrt3\lt 0$, so these are local maxima and are excluded. Thus all positive minimum coordinates are

The two smallest are obtained with $n=0$ and $n=1$:

Prices in the model are in thousands of pounds, so

Taking logarithms gives $T=10\ln2=6.9314718\ldots$, hence $\boxed{T=6.93\text{ years}}$.

Thus the answer is $\boxed{-0.97\text{ thousand pounds per year}}$ to 2 decimal places.

At $t=0$, $36=4+m$, so $\boxed{m=32}$.

The initial car price is $7+16=23$ thousand pounds. The difference is $36-23=13$ thousand pounds, or $\boxed{\pounds13\,000}$.

Equating the models with $m=32$ gives

Let $u=e^{-0.1\alpha}\gt 0$. Then $32u^2-16u-3=0$, so

Reject the negative root because an exponential is positive. Hence

This value is inside the model interval $0\leq t\leq15$.

At $t=0$, $H=32+40-20=52$.

As $t\to\infty$, both decaying exponential terms tend to zero.

At the maximum,

Also $H”(t)=1.6e^{-0.2t}-16.2e^{-0.9t}$. At the stationary point, $e^{-0.9T}=(4/9)e^{-0.2T}$, so $H”(T)=-5.6e^{-0.2T}\lt 0$, confirming a maximum.

For $H=37$,

Taking logarithms gives

Using $-\ln u=\ln(1/u)$ for $u\gt 0$ then gives

Starting with $t_1=10$,

Continuing without rounding intermediate values gives $t_3=10.3954778346\ldots$, $t_4=10.3954789826\ldots$ and $t_5=10.3954789844\ldots$. Thus (i) $t_2=10.3947$, and (ii) $M=10.3955$ minutes, both to 4 decimal places.

Write $y=(16/9)(3x-k)^{-1}$. The chain rule gives

At $x=1$, the given gradient yields

Hence $3-k=\pm2/3$. Both resulting k-values are positive and different from 3.

Since $k\lt 3$, use $k=7/3$. At $x=1$, $y=16/[9(3-7/3)]=8/3$. The tangent gradient is $-12$, so the normal gradient is $1/12$:

With $k=7/3$, the denominator does not vanish on $[1,3]$. Using the reciprocal-linear antiderivative,

Differentiate the velocity:

At a stationary velocity, $e^{t-10}=9e^{-0.75t}$, so $e^{1.75t}=9e^{10}$ and $t=(10+\ln9)/1.75=6.9698426\ldots$ seconds. The second derivative $-e^{t-10}-6.75e^{-0.75t}$ is negative, so this is a maximum. Substitute this unrounded time into v to get $11.88728125\ldots$ metres per second.

Integrating the velocity gives $12t-e^{t-10}+16e^{-0.75t}$. Evaluate between 0 and T:

Rearrange to $12T=116-16e^{-0.75T}+e^{T-10}-e^{-10}$ and divide by 12.

With $T_1=10$, substitution gives $T_2=9.7492587708\ldots$. Continuing with unrounded values gives

Further iteration settles at $9.7293319307\ldots$ seconds. Thus (i) $T_2=9.7493$ and (ii) the race time is $9.7293$ seconds, to four decimal places.

Factor the first denominator:

On the stated domain $x\gt 2$, both denominators and $x+4$ are nonzero. Hence

The quotient rule gives

For every $x\gt 2$, the squared denominator is positive; hence $f'(x)\lt 0$. Therefore f is strictly decreasing on its whole domain.

Set $y=3+2\ln x$. Then $\ln x=(y-3)/2$, so $x=e^{(y-3)/2}$. For $x\ge1$, the original range is $y\ge3$. Interchanging the input and output symbols gives the inverse with its domain.

The equation $g(f(a))=5$ gives $3+2\ln(f(a))=5$, so $\ln(f(a))=1$ and $f(a)=e$. Using the simplified f,

The resulting value is approximately $2.2083\gt 2$, and the inner output $e\ge1$ is admissible for g.

At the start, $t=0$ and $e^{0.3t}=1$. Thus $N(0)=600/(2+1)=200$.

Divide numerator and denominator by $e^{0.3t}$ to write $N=600/(1+2e^{-0.3t})$. As $t\to\infty$, the denominator tends to 1, so the upper limit is 600 fish.

Set $N=500$ and multiply through: $1000+500e^{0.3t}=600e^{0.3t}$, so $e^{0.3t}=10$. Hence $t=\ln10/0.3=7.675283643\ldots$ years. The fractional year is $12(0.675283643\ldots)=8.10340372\ldots$ months. Round only the months to the nearest whole month.

The quotient rule gives

The two $e^{0.6t}$ terms cancel exactly.

Let $z=e^{0.3T}$. Since $T\ge0$, require $z\ge1$. Then $360z/(2+z)^2=8$ gives $45z=(2+z)^2$, hence

The smaller root lies between 0 and 1 and would give $T\lt 0$, so reject it. For the larger root,

With $u=6x^2+4x-2$ and $v=2x+1$, the quotient rule gives

At $x=2$, $f'(2)=80/25=16/5$, so the normal gradient is $-5/16$. Through $P=(2,6)$,

Multiplying the proposed identity by $2x+1$ gives

Compare coefficients: $2A=6$, $A+2B=4$, $B+D=-2$. Thus $f(x)=3x+\frac12-\frac{5}{2(2x+1)}$.

The curve meets the $x$-axis when $6x^2+4x-2=2(3x-1)(x+1)=0$. The root $x=-1$ is outside $x\gt -1/2$, leaving $x=1/3$. The normal meets the axis at $x=106/5$. Split the shaded region at $P$: a curve area from $1/3$ to 2, then a triangle.

The triangle has area $\frac12(106/5-2)\cdot6=288/5$. Adding gives the exact area of $R$.

Factor the denominator as $(x-4)(x+3)$ and multiply the proposed identity through by it:

The right side expands to $x^4-x^3+(P-12)x^2+(Q-P)x+3Q-12P$. Comparing coefficients gives $P-12=-10$, so $P=2$, and $Q-P=3$, so $Q=5$. The constant check is $3(5)-12(2)=-9$, as required. The identity holds where the original fraction is defined: the stated $x\gt -3$ still excludes $x=4$.

From (a), $g(x)=x^2+2+5/(x-4)$ on the curve domain. Thus $g'(x)=2x-5/(x-4)^2$. At $x=2$, $g(2)=4+2-5/2=7/2$ and $g'(2)=4-5/4=11/4$. The tangent through $(2,7/2)$ is

The source shading is above the x-axis from $x=0$ to $x=2$, so integrate g over those bounds.

The absolute value is essential because $x-4\lt 0$ throughout this interval. Evaluating gives

Thus $a=20/3$ and $b=-5$ in the requested form.