Pure Mathematics Practice: Differentiation — Answers

Questions: Differentiation Drills.

Question 1

(a) By the product and chain rules,

$$\frac{dy}{dx}=6x(x^2+1{)}^2\sin x+(x^2+1{)}^3\cos x.$$

(b) At $x=0$,

$$\frac{dy}{dx}=1.$$

Question 2

At $x=2$,

$$y=2^3-3(2)+4=6.$$

Also,

$$\frac{dy}{dx}=3x^2-3,$$

so the tangent gradient at $x=2$ is $9$.

(a)

$$y-6=9(x-2),$$

or

$$y=9x-12.$$

(b) The normal gradient is $-1/9$, so

$$y-6=-\frac{1}{9}(x-2).$$

Question 3

(a) Differentiating implicitly with respect to $x$ gives

$$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0.$$

Hence

$$\frac{dy}{dx}=-\frac{2x+y}{x+2y},$$

provided $x+2y\ne 0$.

(b)

$$\frac{dx}{dt}=2t,\frac{dy}{dt}=3t^2-1.$$

Thus, for $t\ne 0$,

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3t^2-1}{2t}.$$

At $t=2$,

$$\frac{dy}{dx}=\frac{11}{4}.$$

Question 4

(a) If the width is $x$, the length is $20-x$ because the perimeter remains $40$ cm. Therefore

$$A=x(20-x)=20x-x^2.$$

(b)

$$\frac{dA}{dx}=20-2x.$$

The stationary value occurs at $x=10$. Since

$$\frac{d^{2}A}{d{x}^2}=-2<0,$$

it is a maximum. The corresponding rectangle is a $10$ cm by $10$ cm square, with maximum area

$$100{\text{ cm}}^2.$$

(c) By the chain rule,

$$\frac{dA}{dt}=\frac{dA}{dx}\frac{dx}{dt}=(20-2x)\frac{dx}{dt}.$$

At the stated instant,

$$\frac{dA}{dt}=(20-12)(0.2)=1.6{\text{ cm}}^2{\text{ s}}^{-1}.$$

The positive sign means that the area is increasing at this instant.