Mixed differentiation rules

Text

Compute the following derivatives, using where needed the product, quotient or chain rule:

• (a) $\left[\dfrac{x^2}{2}+x^3\right]'$
• (b) $\left[\dfrac{1}{x}\right]'$
• (c) $\big[x^2\sin(x)\big]'$
• (d) $\left[\dfrac{e^x}{x-3}\right]'$
• (e) $\big[(\ln x)^2\big]'$
• (f) $\big[\sqrt{e^x}\big]'$

Solution

• (a) $\left[\dfrac{x^2}{2}+x^3\right]' = x + 3x^2$
• (b) $\left[x^{-1}\right]' = -x^{-2} = -\dfrac{1}{x^2}$
• (c) $\big[x^2\sin x\big]' = 2x\sin x + x^2\cos x$ (product)
• (d) $\left[\dfrac{e^x}{x-3}\right]' = \dfrac{e^x(x-3) - e^x}{(x-3)^2} = \dfrac{e^x(x-4)}{(x-3)^2}$ (quotient)
• (e) $\big[(\ln x)^2\big]' = 2\ln x\cdot\dfrac{1}{x} = \dfrac{2\ln x}{x}$ (chain)
• (f) $\sqrt{e^x}=e^{x/2}\Rightarrow \big[e^{x/2}\big]' = \dfrac{1}{2}e^{x/2} = \dfrac{\sqrt{e^x}}{2}$ $\boxed{x+3x^2;\ \ -\tfrac{1}{x^2};\ \ 2x\sin x+x^2\cos x;\ \ \tfrac{e^x(x-4)}{(x-3)^2};\ \ \tfrac{2\ln x}{x};\ \ \tfrac{\sqrt{e^x}}{2}}$