Pre-interview_Test_Question_1

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For (ai) it’s neater to write:

f(x) = 4x^{-1/4} + x (x^2 - 1)^{-1/3}

Then using the product rule (and chain rule) on the second term:

\displaystyle f'(x) = -x^{-5/4} + (x^2 - 1)^{-1/3} - \frac 1 3 (x^2 - 1)^{-4/3} \times \frac {\mathrm d} {\mathrm dx} (x^2 - 1) \\= -x^{-5/4} + (x^2 - 1)^{-1/3} - \dfrac 2 3 x^2 (x^2 - 1)^{-4/3}

For (aii) we can use the chain rule:

\begin{align*}f'(x) & = \frac {\mathrm d} {\mathrm dx} \left(\sin^2 \frac 1 x\right) \exp\left(\sin^2 \frac 1 x\right) \\ & = \frac 1 2 \frac {\mathrm d} {\mathrm dx} \left(1 - \cos \frac 2 x\right) \exp\left(\sin^2 \frac 1 x\right) \\ & = \frac 1 2 \times \frac {\mathrm d} {\mathrm dx} \left(\frac 1 x\right)\left(-2 \sin \frac 2 x\right) \exp\left(\sin^2 \frac 1 x\right)\\ & = -\frac 1 {x^2} \sin \left(\frac 2 x\right) \exp\left(\sin^2 \frac 1 x\right)\end{align*}

(bi) Is a standard A-level integral, by parts is the way to go:

\begin{align*}\int x \sin x \mathrm dx &= \int x (-\cos x)' \mathrm dx \\ &= -x \cos x - \int (-\cos x) \mathrm dx \\ & = -x \cos x + \int \cos x \mathrm dx \\ & = -x \cos x + \sin x + C\end{align*}

(bii) Another standard one, let \theta = \arcsin x so that x = \sin \theta and \mathrm dx = \cos \theta \mathrm d\theta. Then:

\begin{align*}\int \frac {\mathrm dx} {\sqrt {1 - x^2}} & = \int \frac {\cos \theta} {\sqrt {1 - \sin^2 \theta}} \mathrm d\theta \\ & = \int \frac {\cos \theta} {\sqrt {\cos^2 \theta}} \mathrm d\theta \\ & = \int \mathrm d\theta\\ & = \theta + C \\ & = \arcsin x + C\end{align*}

(writing \sqrt {\cos^2 \theta} = \cos \theta since \cos \theta > 0 on -\dfrac \pi 2 < \theta < \dfrac \pi 2)

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bii my new favourite integral substitution haha