Nine derivatives with the chain rule

Text

Compute the derivative of the following functions, indicating the composition scheme when the function is composite.

i) $f(x)=2x^4+3x^2+4x+6$; ii) $f(x)=e^x\sin x$; iii) $f(x)=e^{2+2x^2}$; iv) $f(x)=\dfrac1{\sin x}$; v) $f(x)=\sqrt[3]{x^2+\cos x}$; vi) $f(x)=\dfrac1{x^3}$; vii) $f(x)=\tan\!\big(\ln\sqrt{xe^x}\big)$; viii) $f(x)=\dfrac1{\ln\sqrt x}$; ix) $f(x)=\dfrac{e^x}{x}$.

Solution

i) Polynomial, term-by-term derivative: $f'(x)=8x^3+6x+4.$

ii) Product rule: $f'(x)=e^x\sin x+e^x\cos x=e^x(\sin x+\cos x).$

iii) Composite $e^{u}$ with $u=2+2x^2$, $u'=4x$: $f'(x)=e^{2+2x^2}\cdot 4x=4x\,e^{2+2x^2}.$

iv) $\dfrac1{\sin x}=(\sin x)^{-1}$, composite with $u=\sin x$: $f'(x)=-(\sin x)^{-2}\cos x=-\frac{\cos x}{\sin^2 x}.$

v) $\sqrt[3]{u}=u^{1/3}$ with $u=x^2+\cos x$, $u'=2x-\sin x$: $f'(x)=\tfrac13 u^{-2/3}\,u'=\frac{2x-\sin x}{3\sqrt[3]{(x^2+\cos x)^2}}.$

vi) $\dfrac1{x^3}=x^{-3}$: $f'(x)=-3x^{-4}=-\frac{3}{x^4}.$

vii) First simplify the argument: $\ln\sqrt{xe^x}=\tfrac12\ln(xe^x)=\tfrac12(\ln x+x).$ Hence $f(x)=\tan(u)$ with $u=\tfrac12(\ln x+x)$, $u'=\tfrac12\!\left(\dfrac1x+1\right)=\dfrac{x+1}{2x}$: $f'(x)=\sec^2(u)\,u'=\frac{x+1}{2x}\,\sec^2\!\Big(\tfrac12(x+\ln x)\Big).$

viii) $\ln\sqrt x=\tfrac12\ln x$, so $f(x)=\dfrac{2}{\ln x}=2(\ln x)^{-1}$: $f'(x)=-2(\ln x)^{-2}\cdot\frac1x=-\frac{2}{x\ln^2 x}.$

ix) Quotient rule: $f'(x)=\frac{e^x\cdot x-e^x\cdot 1}{x^2}=\frac{e^x(x-1)}{x^2}.$

The most involved result is the one in part vii): $\boxed{\;f'(x)=\dfrac{x+1}{2x}\,\sec^2\!\Big(\tfrac12(x+\ln x)\Big)\;}$