# How do you differentiate f(x)= ( x + 1 )/ ( x - 4csc x ) using the quotient rule?

Mar 16, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{4 x \csc x \cot x + 4 \csc x \cot x - 4 \csc x - 1}{x - 4 \csc x} ^ 2$

#### Explanation:

According to quotient rule

d/dx((f(x))/(g(x)))=(d/dx(f(x))xxg(x)-f(x)xxd/dx(g(x)))/((g(x))^2

Hence as $f \left(x\right) = \frac{x + 1}{x - 4 \csc x}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{d}{\mathrm{dx}} \left(x + 1\right) \times \left(x - 4 \csc x\right) - \left(x + 1\right) \times \frac{d}{\mathrm{dx}} \left(x - 4 \csc x\right)}{x - 4 \csc x} ^ 2$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1 \times \left(x - 4 \csc x\right) - \left(x + 1\right) \times \left(1 - 4 \left(- \csc x \cot x\right)\right)}{x - 4 \csc x} ^ 2$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{x - 4 \csc x - x - 1 + 4 \left(x + 1\right) \csc x \cot x}{x - 4 \csc x} ^ 2$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{x - 4 \csc x - x - 1 + 4 x \csc x \cot x + 4 \csc x \cot x}{x - 4 \csc x} ^ 2$ or
$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{4 x \csc x \cot x + 4 \csc x \cot x - 4 \csc x - 1}{x - 4 \csc x} ^ 2$