202211220908 Solution to 1999-CE-AMATH-I-1

Find

(a) $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\sin (x^2+1)}$ ,
(b) $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac{\sin (x^2+1)}{x}\bigg]}$ .

Roughwork.

Let $g(x)=\sin x$ and $h(x)=x^2+1$ s.t.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\sin (x^2+1) & = \frac{\mathrm{d}}{\mathrm{d}x}(g\circ h)(x) \\ & = (g\circ h)'(x) \\ & = (g'\circ h)\cdot h'(x) \\ & = \frac{\mathrm{d}g(h(x))}{\mathrm{d}h(x)}\cdot\frac{\mathrm{d}h(x)}{\mathrm{d}x} \\ & = \cos (h(x))\cdot 2x \\ & = 2x\cos (x^2+1) \end{aligned}$

Next, let $f(x)=x^{-1}$ s.t.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac{\sin (x^2+1)}{x}\bigg] & = \frac{\mathrm{d}}{\mathrm{d}x}\Big( g(h(x))\cdot f(x)\Big) \end{aligned}$

The remaining are left the reader as an exercise.

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