# If f(x)= cot2 x  and g(x) = e^(-3x ) , how do you differentiate f(g(x))  using the chain rule?

$\textcolor{red}{\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = 6 \cdot {e}^{- 3 x} {\csc}^{2} \left(2 \cdot {e}^{- 3 x}\right)}$

#### Explanation:

$f \left(g \left(x\right)\right) = \cot \left(2 \cdot g \left(x\right)\right)$

$f \left(g \left(x\right)\right) = \cot \left(2 \cdot {e}^{- 3 x}\right)$

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \cot \left(2 \cdot {e}^{- 3 x}\right)$

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = - {\csc}^{2} \left(2 \cdot {e}^{- 3 x}\right) \cdot \frac{d}{\mathrm{dx}} \left(2 \cdot {e}^{- 3 x}\right)$

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = - {\csc}^{2} \left(2 \cdot {e}^{- 3 x}\right) \cdot 2 \cdot \frac{d}{\mathrm{dx}} \left({e}^{- 3 x}\right)$

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = - {\csc}^{2} \left(2 \cdot {e}^{- 3 x}\right) \cdot 2 \cdot {e}^{- 3 x} \cdot \frac{d}{\mathrm{dx}} \left(- 3 x\right)$

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = - {\csc}^{2} \left(2 \cdot {e}^{- 3 x}\right) \cdot 2 \cdot {e}^{- 3 x} \cdot \left(- 3\right)$

$\textcolor{red}{\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = 6 \cdot {e}^{- 3 x} {\csc}^{2} \left(2 \cdot {e}^{- 3 x}\right)}$

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