Differentiation review: Differentiate each of the functions that follow. Simplify your answers as much as possible.

a. $f\left(x\right)=x{e}^x-e^x$

b. $f\left(x\right)=\frac{x^2\ln x}{2}-\frac{x^2}{4}$

c. $f\left(x\right)=-e^{-x}(x^2+1)-2x{e}^{-x}-2e^{-x}$

d.$f\left(x\right)=\frac{x^3}{3}-2(x{e}^x-e^x)+\frac{e^{2x}}{2}$

e. $f\left(x\right)=-x\cot x+\ln |\sin x|$

f. $f\left(x\right)=x^3\sin x+3x^2\cos x-6x\sin x-6\cos x$

Given function, $f\left(x\right)=x{e}^x-e^x$

The derivative, $f\text{'}\left(x\right)=x\left(e^x\right)+e^x\left(1\right)-e^x$

$\Rightarrow f\text{'}\left(x\right)=x{e}^x$

Given function,$f\left(x\right)=\frac{x^2\ln x}{2}-\frac{x^2}{4}$

The derivative, $f\text{'}\left(x\right)=\frac{x^2\left({\displaystyle \frac{1}{x}}\right)+\ln x\left(2x\right)}{2}-\frac{2x}{4}$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{x}{2}+x\ln x-\frac{x}{2} \\ \Rightarrow f\text{'}\left(x\right)=x\ln x \end{gathered}$$

Given function, $f\left(x\right)=-e^{-x}(x^2+1)-2x{e}^{-x}-2e^{-x}$

$\Rightarrow f\left(x\right)=-e^{-x}{x}^2-e^{-x}-2x{e}^{-x}-2e^{-x}$

The derivative, role="math" localid="1653980597687" $f\text{'}\left(x\right)=-\left[e^{-x}\left(2x\right)+x^2(-e^{-x})\right]-\left(-e^{-x}\right)-2\left[x\left(-e^{-x}\right)+e^{-x}\left(1\right)\right]-2\left(-e^{-x}\right)$

role="math" localid="1653982097270" $$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=-2x{e}^{-x}+x^2{e}^{-x}+e^{-x}+2x{e}^{-x}-2e^{-x}+2e^{-x} \\ \Rightarrow f\text{'}\left(x\right)=x^2{e}^{-x}+e^{-x} \end{gathered}$$

Given function,$f\left(x\right)=\frac{x^3}{3}-2(x{e}^x-e^x)+\frac{e^{2x}}{2}$

The derivative, $f\text{'}\left(x\right)=\frac{3x^2}{3}-2\left[\left(x{e}^x+e^x\left(1\right)\right)-e^x\right]+\frac{2e^{2x}}{2}$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=x^2-2\left[x{e}^x+e^x-e^x\right]+e^{2x} \\ \Rightarrow f\text{'}\left(x\right)=x^2-2x{e}^x+e^{2x} \end{gathered}$$

Given function,$f\left(x\right)=-x\cot x+\ln |\sin x|$

The derivative, $f\text{'}\left(x\right)=-x(-{\csc }^{2}x)-\cot x\left(1\right)+\frac{1}{\sin x}\times \cos x$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=x{\csc }^{2}x-\cot x+\cot x \\ \Rightarrow f\text{'}\left(x\right)=x{\csc }^{2}x \end{gathered}$$

Given function,$f\left(x\right)=x^3\sin x+3x^2\cos x-6x\sin x-6\cos x$

The derivative, $f\text{'}\left(x\right)=x^3\cos x+\sin x\left(3x^2\right)+3\left[x^2(-\sin x)+\cos x\left(2x\right)\right]-6\left[x\cos x+\sin x\left(1\right)\right]-6(-\sin x)$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=x^3\cos x+3x^2\sin x-3x^2\sin x+6x\cos x-6x\cos x-6\sin x+6\sin x \\ \Rightarrow f\text{'}\left(x\right)=x^3\cos x \end{gathered}$$