# How do you differentiate f(x)= ( 4- cscx )/ (e^x + 2)  using the quotient rule?

$\textcolor{red}{f ' \left(x\right) = \frac{\left({e}^{x} \cdot \csc x \cdot \cot x + 2 \cdot \csc x \cdot \cot x - 4 \cdot {e}^{x} + {e}^{x} \cdot \csc x\right)}{{\left({e}^{x} + 2\right)}^{2}}}$

#### Explanation:

Start from the given function

$f \left(x\right) = \frac{4 - \csc x}{{e}^{x} + 2}$

The formula to be used in finding the derivative is

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

Let $u = \left(4 - \csc x\right)$ and $v = \left({e}^{x} + 2\right)$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{4 - \csc x}{{e}^{x} + 2}\right) = \frac{\left({e}^{x} + 2\right) \frac{d}{\mathrm{dx}} \left(4 - \csc x\right) - \left(4 - \csc x\right) \frac{d}{\mathrm{dx}} \left({e}^{x} + 2\right)}{{\left({e}^{x} + 2\right)}^{2}}$

$f ' \left(x\right) = \frac{\left({e}^{x} + 2\right) \left(0 - \left(- \csc x \cdot \cot x\right)\right) - \left(4 - \csc x\right) \left({e}^{x} \cdot 1 + 0\right)}{{\left({e}^{x} + 2\right)}^{2}}$

$\textcolor{red}{f ' \left(x\right) = \frac{\left({e}^{x} \cdot \csc x \cdot \cot x + 2 \cdot \csc x \cdot \cot x - 4 \cdot {e}^{x} + {e}^{x} \cdot \csc x\right)}{{\left({e}^{x} + 2\right)}^{2}}}$

God bless....I hope the explanation is useful.