CAIE IGCSE Additional Math 14 calculus

Use this Calculus hub to move between differentiation, curve applications, connected rates, optimisation and reverse differentiation.

Covered syllabus points

14.1. Understand derived functions 14.2. Use differentiation notation 14.3. Differentiate standard functions 14.4. Differentiate products and quotients 14.5. Find gradients, tangents and normals 14.6. Find stationary points 14.7. Apply differentiation to rates and approximations 14.8. Apply differentiation to maxima and minima 14.9. Use derivative tests for maxima and minima 14.10. Understand integration as reverse differentiation 14.11. Integrate powers of x 14.12. Integrate standard composite functions 14.13. Evaluate definite integrals and areas 14.14. Apply calculus to kinematics 14.15. Use kinematics graphs

[Maximum number: 2]

It is given that $y=3\tan^2x$ for $0^\circ<x<360^\circ$ .

Show that $\frac{dy}{dx}=m\tan x\sec^2x$ , where m is an integer to be found.

$\frac{dy}{dx}=6\tan x\sec^2x$
B2; B1 for $\frac{d}{dx}(\tan^2x)=2\tan x\sec^2x$

[Maximum number: 10]

The equation of a curve is $y=kxe^{-2x}$ , where k is a constant.

(a)

Find $\frac{dy}{dx}$ .

[ 2 ]

$\frac{d}{dx}(e^{-2x})=-2e^{-2x}$
B1
$\frac{dy}{dx}=ke^{-2x}-2kxe^{-2x}$
B1
FT for product rule $k e^{-2x}+kx\cdot\frac{d}{dx}(e^{-2x})$ .

(b)

Find the coordinates of the stationary point on the curve $y=10xe^{-2x}$ .

[ 3 ]

Equate $\frac{dy}{dx}=0$ and find
10-20x=0
M1
$\left(\frac{1}{2},\frac{5}{e}\right)$
A2; A1 for $x=\frac{1}{2}$ only. Equivalent $y=5e^{-1}$ or decimal 1.84 accepted.

(c)

Use your answer to part (a) to find $\int 4xe^{-2x}\,dx$ .

[ 3 ]

$-2xe^{-2x}-e^{-2x}+c$
B3
B2 for $-2xe^{-2x}-e^{-2x}$ , or $\int4xe^{-2x}\,dx=-2xe^{-2x}+\int2e^{-2x}\,dx$ .
B1 for using $kxe^{-2x}=\int(ke^{-2x}-2kxe^{-2x})\,dx$ or better.

(d)

Find the exact value of $\int_0^1 4xe^{-2x}\,dx$ .

[ 2 ]

$-2e^{-2}-e^{-2}-(0-e^0)$
M1
$1-\frac{3}{e^2}$
or $1-3e^{-2}$
A1

[Maximum number: 7]

A curve has equation $y=\ln(5-3x)$ , where $x<\frac{5}{3}$ . The normal to the curve at the point where x=-5 cuts the x-axis at the point P.
Find the equation of the normal and the x-coordinate of P.

$\frac{dy}{dx}=\frac{-3}{5-3x}, \qquad \left.\frac{dy}{dx}\right|_{x=-5}=-\frac{3}{20}.$
At (x=-5), (y=\ln20), so the normal gradient is $\frac{20}{3}$ . Hence
$y-\ln20=\frac{20}{3}(x+5).$
At the (x)-axis, (y=0), giving
$x=-5-\frac{3\ln20}{20}=-5.449\ldots.$

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Covered syllabus points

14. Calculus question 1

14. Calculus question 2

Question (a)

Question (b)

Question (c)

Question (d)

14. Calculus question 3