CAIE A-Level Mathematics AS 1.7 Differentiation Questions

Practise differentiating powers and composite expressions and applying derivatives to gradients, tangents, normals, stationary points, monotonicity and connected rates.

Exam points

apply power and chain rules and simplify first or second derivatives into usable form
set dy/dx to a gradient or zero to find tangents, normals and stationary points
connect rates with dy/dt = (dy/dx)(dx/dt) and substitute values with consistent units

Covered syllabus points

1.7.1—Gradient as limit 1.7.2—Differentiation rules 1.7.3—Differentiation applications 1.7.4—Stationary points

[Maximum number: 3]

The equation of a curve is y=f(x), where $\mathrm{f}(x)=(2 x-1) \sqrt{3 x-2}-2$ . The following points lie on the curve. Non-exact values have been given correct to 5 decimal places.

A(2,4), B(2.0001, k), C(2.001,4.00625), D(2.01,4.06261), E(2.1,4.63566), F(3,11.22876)

(a)

Find the value of k. Give your answer correct to 5 decimal places.

The table shows the gradients of the chords A B, A C, A D and A F.

Table for Question (a) — CAIE A-Level Mathematics AS

[ 1 ]

(b)

Find the gradient of the chord A E. Give your answer correct to 4 decimal places.

[ 1 ]

(c)

Deduce the value of $f^{\prime}(2)$ using the values in the table.

[ 1 ]

Suggests that $[\mathrm{f}(2)]=6.25$
CAO

[Maximum number: 9]

The equation of a curve is $y=(3-2 x)^{3}+24 x$ .

(a)

Find expressions for $\frac{\mathrm{d} y}{\mathrm{~d} x}$ and $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}$ .

[ 4 ]

$\frac{dy}{dx}=3(3-2x)^2(-2)+24=-6(3-2x)^2+24.$
B1, B1

$\frac{d^2y}{dx^2}=-12(3-2x)(-2)=24(3-2x).$
B1FT, B1

(b)

Find the coordinates of each of the stationary points on the curve.

[ 3 ]

Stationary points occur when
$\frac{dy}{dx}=0.$ $6(3-2x)^2=24,$
so
$3-2x=\pm2.$ $x=\frac12,\quad y=20,$
or
$x=2\frac12,\quad y=52.$

(c)

Determine the nature of each stationary point.

[ 2 ]

At $x=\frac12$ ,
$\frac{d^2y}{dx^2}=48,$
so the stationary point is a minimum.
B1FT

At $x=2\frac12$ ,
$\frac{d^2y}{dx^2}=-48,$
so the stationary point is a maximum.
B1FT

[Maximum number: 6]

Figure for Question 3 — CAIE A-Level Mathematics AS

The diagram shows the curve with equation $y=9\left(x^{-\frac{1}{2}}-4 x^{-\frac{3}{2}}\right)$ . The curve crosses the x-axis at the point A.

(a)

Find the equation of the tangent to the curve at A.

[ 4 ]

$\frac{\mathrm{d} y}{\mathrm{~d} x}=9\left(-\frac{1}{2} x^{-\frac{3}{2}}+6 x^{-\frac{5}{2}}\right)$
B2, 1, 0
all 3 terms correct: $9,-\frac{1}{2} x^{-\frac{3}{2}}$ and $6 x^{-\frac{5}{2}}$
2 of the 3 terms correct
At x=4 gradient $=9\left(-\frac{1}{16}+\frac{6}{32}\right)=\frac{9}{8}$
Using their x=4 in their differentiated expression
and attempt to find equation of the tangent.
Equation is $y=\frac{9}{8}(x-4)$
or $y=\frac{9 x}{8}-\frac{9}{2} \mathrm{OE}$

(b)

Find the x-coordinate of the maximum point of the curve.

[ 2 ]

$9 x^{-\frac{5}{2}}\left(-\frac{1}{2} x+6\right)=0$
Set their $\frac{\mathrm{d} y}{\mathrm{~d} x}$ to zero and an attempt to solve.
x=12
Condone $( \pm) 12$ from use of a correct method.

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)