WASSCE Further Maths: Calculus (Differentiation & Integration) Practice Questions & Answers

Using the chain rule (function of a function), let $u = 3x^2 - 5$u=3x2−5, meaning $y = u^4$y=u4. We find $\frac{dy}{du} = 4u^3$dudy​=4u3 and $\frac{du}{dx} = 6x$dxdu​=6x. Multiplying these gives $\frac{dy}{dx} = 4(3x^2 - 5)^3 \times 6x = 24x(3x^2 - 5)^3$dxdy​=4(3x2−5)3×6x=24x(3x2−5)3.

Differentiate both sides with respect to $x$x to get $2x + 2y \frac{dy}{dx} = 0$2x+2ydxdy​=0. Rearranging to solve for $\frac{dy}{dx}$dxdy​ gives $2y \frac{dy}{dx} = -2x$2ydxdy​=−2x, which simplifies to $\frac{dy}{dx} = -\frac{x}{y}$dxdy​=−yx​.

The volume of a sphere is $V = \frac{4}{3}\pi r^3$V=34​πr3. Differentiating gives $\frac{dV}{dr} = 4\pi r^2$drdV​=4πr2. Using the chain rule, $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$dtdV​=drdV​×dtdr​. Given $\frac{dV}{dt} = -10$dtdV​=−10 (decreasing) and $r = 5$r=5, we have $-10 = 4\pi(5)^2 \times \frac{dr}{dt} \implies -10 = 100\pi \times \frac{dr}{dt}$−10=4π(5)2×dtdr​⟹−10=100π×dtdr​. Thus, $\frac{dr}{dt} = -\frac{1}{10\pi} \text{ cm/s}$dtdr​=−10π1​ cm/s.

Integrate each term using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$∫xndx=n+1xn+1​+C. $\int 4x^3 dx = x^4$∫4x3dx=x4 and $\int -6x dx = -3x^2$∫−6xdx=−3x2. Adding the constant of integration yields $x^4 - 3x^2 + C$x4−3x2+C.

Using the quotient rule $\frac{v u' - u v'}{v^2}$v2vu′−uv′​, let $u = 2x$u=2x and $v = x-1$v=x−1. Then $u' = 2$u′=2 and $v' = 1$v′=1. The derivative is $\frac{(x-1)(2) - (2x)(1)}{(x-1)^2} = \frac{2x - 2 - 2x}{(x-1)^2} = \frac{-2}{(x-1)^2}$(x−1)2(x−1)(2)−(2x)(1)​=(x−1)22x−2−2x​=(x−1)2−2​.

Let $u = x^2 + 1$u=x2+1. Then $du = 2x dx$du=2xdx, which implies $x dx = \frac{1}{2} du$xdx=21​du. Substituting gives $\int \frac{1}{2} u^{1/2} du$∫21​u1/2du. Integrating yields $\frac{1}{2} \left( \frac{u^{3/2}}{3/2} \right) + C = \frac{1}{3}u^{3/2} + C$21​(3/2u3/2​)+C=31​u3/2+C. Substituting back $u$u results in $\frac{1}{3}(x^2+1)^{3/2} + C$31​(x2+1)3/2+C.

The anti-derivative of $3x^2$3x2 is $x^3$x3. Evaluating this from $x = 1$x=1 to $x = 2$x=2 gives $[2^3] - [1^3] = 8 - 1 = 7$[23]−[13]=8−1=7.

The gradient is given by $\frac{dy}{dx}$dxdy​. Differentiating gives $\frac{dy}{dx} = 3x^2 - 4x + 1$dxdy​=3x2−4x+1. Substituting $x = 2$x=2, we get $3(2)^2 - 4(2) + 1 = 12 - 8 + 1 = 5$3(2)2−4(2)+1=12−8+1=5.

Find turning points where $\frac{dy}{dx} = 0$dxdy​=0. $\frac{dy}{dx} = x^2 - 2x - 3 = (x-3)(x+1) = 0$dxdy​=x2−2x−3=(x−3)(x+1)=0, so $x=3$x=3 and $x=-1$x=−1. The second derivative is $\frac{d^2y}{dx^2} = 2x - 2$dx2d2y​=2x−2. At $x=3$x=3, $2(3)-2 = 4 > 0$2(3)−2=4>0 (Minimum). At $x=-1$x=−1, $2(-1)-2 = -4 < 0$2(−1)−2=−4<0 (Maximum).

The derivative of $\sin x$sinx is $\cos x$cosx, and the derivative of $\cos x$cosx is $-\sin x$−sinx. Therefore, $\frac{dy}{dx} = \cos x - \sin x$dxdy​=cosx−sinx.