# Question #4492c

Jan 7, 2018

A) $\frac{d}{\mathrm{dx}} \left({e}^{x} \sin x\right) = {e}^{x} \left(\cos x + \sin x\right)$

B) $\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right) = \frac{2}{x + 1} ^ 2$

C) $\frac{d}{\mathrm{dx}} \left(\frac{1}{1 - {x}^{2}}\right) = \frac{2 x}{1 - {x}^{2}} ^ 2$

#### Explanation:

A) Using the product rule:

$\frac{d}{\mathrm{dx}} \left({e}^{x} \sin x\right) = {e}^{x} \frac{d}{\mathrm{dx}} \sin x + \sin x \frac{d}{\mathrm{dx}} {e}^{x}$

$\frac{d}{\mathrm{dx}} \left({e}^{x} \sin x\right) = {e}^{x} \cos x + {e}^{x} \sin x$

$\frac{d}{\mathrm{dx}} \left({e}^{x} \sin x\right) = {e}^{x} \left(\cos x + \sin x\right)$

B) Using the quotient rule:

$\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right) = \frac{\left(x + 1\right) \frac{d}{\mathrm{dx}} \left(x - 1\right) - \left(x - 1\right) \frac{d}{\mathrm{dx}} \left(x + 1\right)}{x + 1} ^ 2$

$\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right) = \frac{x + 1 - x + 1}{x + 1} ^ 2$

$\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right) = \frac{2}{x + 1} ^ 2$

C) Using the chain rule:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{1 - {x}^{2}}\right) = - \frac{1}{1 - {x}^{2}} ^ 2 \frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right)$

$\frac{d}{\mathrm{dx}} \left(\frac{1}{1 - {x}^{2}}\right) = \frac{2 x}{1 - {x}^{2}} ^ 2 = 2 x {y}^{2}$