TB. 1 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Part (b) $f\text{'}\left(x\right)=x \ln x$

Part (c) $f\text{'}\left(x\right)=x^2 e^{-x}+e^{-x}$

Part (d) $f\text{'}\left(x\right)=x^2- 2x{e}^x+ e^{2x}$

Part (e) $f\text{'}\left(x\right)=x {\csc }^{2}x$

Part (f) $f\text{'}\left(x\right)=x^3 \cos x$

Part (a) Step 1. Finding the derivative

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

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

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

Part (b) Step 1. Finding the derivative

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

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

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

Part (c) Step 1. Finding the derivative

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

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

Part (d) Step 1. Finding the derivative

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

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

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

Part (e) Step 1. Finding the derivative

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

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

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

Part (f) Step 1. Finding the derivative

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

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

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