Question 10.10 • Calculus (4th Edition) • Textbookish

We have

\begin{align*} f'(x)&=2x\sin\frac{1}{x}+x^2\cos\frac{1}{x}\left(-\frac{1}{x^2}\right) \\ &=2x\sin\frac{1}{x}-\cos\frac{1}{x} \end{align*}

and

f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}h\sin\frac{1}{h}=0.

Hence,

\begin{align*} \text{i)}\quad (f\circ h)'(0)&=f'(h(0))h'(0) \\ &=\left(6\sin\frac{1}{3}-\cos\frac{1}{3}\right)\sin^2(\sin 1), \end{align*}

\begin{align*} \text{ii)}\quad (k\circ f)'(0)&=k'(f(0))f'(0) \\ &=\sin 1\cdot 0=0. \end{align*}

\begin{align*} \text{iii)}\quad a'(x)&=h'(x^2)\cdot 2x \\ &=\sin^2(\sin(x^2+1))\cdot 2x. \end{align*}

Thus,

a'(x^2)=\sin^2(\sin(x^4+1))\cdot 2x^2.

Solutions