Pearson Edexcel IAL Mathematics P3.4 Differentiation Question Bank

Practise differentiation across trigonometric, exponential, logarithmic, implicit and parametric forms to find gradients, rates and points.

Exam points

differentiate trig or exponential functions to find tangent gradients at specified points
use product, quotient or chain rules on model functions involving e^{kt}
set dy/dx equal to a given gradient to find coordinates or model times

Covered syllabus points

P3.4.1 - Differentiating exp, log and trig functions P3.4.2 - Differentiation P3.4.3 - Reciprocal derivative relationship P3.4.4 - exponential

[Maximum number: 7]

A curve C has equation

y=x^{\sin x} \quad x>0 \quad y>0

(a)

Find, by firstly taking natural logarithms, an expression for $\frac{\mathrm{d} y}{\mathrm{~d} x}$ in terms of x and y.

[ 5 ]

$y=x^{\sin x}\Rightarrow \ln y=\sin x\ln x$

B1

$\frac1y\frac{dy}{dx}=\frac{\sin x}{x}+\ln x\cos x$

M1 M1 A1

$\frac{dy}{dx}=y\left(\frac{\sin x}{x}+\ln x\cos x\right)$

A1

(b)

Hence show that the x coordinates of the stationary points of C are solutions of the equation

\tan x+x \ln x=0

[ 2 ]

$\frac{dy}{dx}=0\Rightarrow y\left(\frac{\sin x}{x}+\ln x\cos x\right)=0$ $\frac{\sin x}{x}+\ln x\cos x=0$

M1

$\frac{\sin x}{\cos x}+x\ln x=0 \Rightarrow \tan x+x\ln x=0$

A1*

[Maximum number: 8]

A scientist is studying a population of fish in a lake. The number of fish, N, in the population, t years after the start of the study, is modelled by the equation

N=\frac{600 \mathrm{e}^{0.3 t}}{2+\mathrm{e}^{0.3 t}} \quad t \geqslant 0

Use the equation of the model to answer parts (a), (b), (c), (d) and (e).

(a)

Find the upper limit to the number of fish in the lake.

[ 1 ]

(b)

Show that

$\frac{dN}{dt}=\frac{Ae^{0.3t}}{(2+e^{0.3t})^2}$

where A is a constant to be found.

[ 3 ]

$N=\frac{600e^{0.3t}}{2+e^{0.3t}}$ $\frac{dN}{dt}=\frac{(2+e^{0.3t})(180e^{0.3t})-600e^{0.3t}(0.3e^{0.3t})}{(2+e^{0.3t})^2}$
M1 A1
$\frac{dN}{dt}=\frac{360e^{0.3t}}{(2+e^{0.3t})^2}$
A1

(c)

Given that when t=T, $\frac{dN}{dt}=8$ ,
find the value of T to one decimal place.

(Solutions relying entirely on calculator technology are not acceptable.)

[ 4 ]

$8=\frac{360e^{0.3t}}{(2+e^{0.3t})^2}$ $8(e^{0.3t})^2-328(e^{0.3t})+32=0$ $e^{0.3t}=\frac{41+\sqrt{1665}}{2}$ $t=\frac{\ln\left(\frac{41+\sqrt{1665}}{2}\right)}{0.3}$ $T\approx12.4$

M1 dM1 ddM1 A1

Marking guidance:
M1 for setting dN/dt=8 and obtaining a quadratic in $e^{0.3t}$ .
dM1 for a correct attempt to solve the quadratic in $e^{0.3t}$ .
ddM1 for correct log work for $e^{0.3t}=k$ , k>0.
A1 for awrt 12.4.

[Maximum number: 6]

A curve C has equation y=f(x), where

f(x)=\arcsin\left(\frac{x}{2}\right),\qquad -2\leqslant x\leqslant2,\qquad -\frac{\pi}{2}\leqslant y\leqslant\frac{\pi}{2}

(a)

Given $x=2\sin y$ , show that

\frac{dy}{dx}=\frac{1}{\sqrt{A-x^2}}

where A is a constant to be found.

[ 3 ]

x=2\sin y\Rightarrow \frac{dx}{dy}=2\cos y

\frac{dy}{dx}=\frac{1}{2\cos y}

Since $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-\frac{x^2}{4}}$ ,

\frac{dy}{dx}=\frac{1}{2\sqrt{1-\frac{x^2}{4}}}=\frac{1}{\sqrt{4-x^2}}

Thus A=4.

M1 A1 A1cso

(b)

The point P lies on C and has y coordinate $\frac{\pi}{4}$ .

Find the equation of the tangent to C at P. Write your answer in the form y=mx+c, where m and c are constants to be found.

[ 3 ]

At P, $y=\frac{\pi}{4}$ , so

x=2\sin\frac{\pi}{4}=\sqrt2

\left.\frac{dy}{dx}\right|_{x=\sqrt2}=\frac{1}{\sqrt{4-2}}=\frac{\sqrt2}{2}

y-\frac{\pi}{4}=\frac{\sqrt2}{2}(x-\sqrt2)

y=\frac{\sqrt2}{2}x-1+\frac{\pi}{4}

M1 dM1 A1

All question bank results loaded

Covered syllabus points

P3.4 - Differentiation question 1

Question (a)

Question (b)

P3.4 - Differentiation question 2

Question (a)

Question (b)

Question (c)

P3.4 - Differentiation question 3

Question (a)

Question (b)