Find the derivative w.r.t. $~~x~$
$~9(i)~~\{f(x)\}^n$
Sol. let $~~~~y=\{f(x)\}^n \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[f(x)]^n\\=n[f(x)]^{n-1}\cdot \frac{d}{dx}[f(x)]\\=n[f(x)]^{n-1}\cdot f'(x)~~\text{(ans.)}$
$~9(ii)~~e^{\phi(x)}$
Sol. let $~~y=e^{\phi(x)}\\ \therefore~~\frac{dy}{dx}\\=\frac{d}{dx}[e^{\phi(x)}]\\=e^{\phi(x)}\cdot \frac{d}{dx}[\phi(x)]\\=e^{\phi(x)}\cdot \phi'(x)~~\text{(ans.)}$
$~(iii)~~\log[\sqrt{f(x)}]$
Sol. let $~~y=\log[\sqrt{f(x)}] \\ \text{or,}~~y=\log[f(x)]^{1/2}=\frac 12\log f(x)\\ \therefore~~\frac{dy}{dx}\\=\frac 12\frac{d}{dx}[\log(f(x))]\\=\frac 12 \cdot \frac{1}{f(x)}\cdot \frac{d}{dx}[f(x)]\\=\frac 12\cdot \frac{1}{f(x)}\cdot f'(x)\\=\frac{f'(x)}{2f(x)}~~\text{(ans.)}$
$~(iv)~~~3^{u(x)}$
Sol. let $~y=3^{u(x)} \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[3^{u(x)}]\\=3^{u(x)}\cdot \log_e3 \cdot \frac{d}{dx}[u(x)]\\=3^{u(x)}\cdot \log_e3\cdot u'(x)~~\text{(ans.)}$
$~(v)~~\sin\{f(x)\}$
Sol. let $~~y=\sin[f(x)] \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\sin(f(x))]\\=\cos(f(x))\cdot \frac{d}{dx}[f(x)]\\=\cos(f(x))\cdot f'(x)~~\text{(ans.)}$
$~(vi)~~\sec \phi(x)$
Sol. let $~y=\sec \phi(x) \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[\sec~\phi(x)]\\=\sec~\phi(x)\cdot \tan~\phi(x)~\cdot \frac{d}{dx}[\phi(x)]\\=\sec~\phi(x)\tan~\phi(x) \cdot ~\phi'(x)~~\text{(ans.)}$
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$~(vii)~~\cos^{-1}~[f(x)]$
Sol. let $~~y=\cos^{-1}~[f(x)] \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\cos^{-1}~f(x)]\\=\frac{-1}{\sqrt{1-(f(x))^2}} \cdot \frac{d}{dx}(f(x))\\=\frac{-1}{\sqrt{1-(f(x))^2}} \cdot f'(x)\\=\frac{-f'(x)}{\sqrt{1-(f(x))^2}}~~\text{(ans.)}$
$~(viii)~~\tan^{-1}[\phi(x)]$
Sol. let $~~y=\tan^{-1}[\phi(x)] \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[\tan^{-1}[\phi(x)]]\\=\frac{1}{1+[\phi(x)]^2} \cdot \frac{d}{dx}[\phi(x)]\\=\frac{1}{1+[\phi(x)]^2} \cdot \phi'(x)\\=\frac{\phi'(x)}{1+[\phi(x)]^2}~~\text{(ans.)}$
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