First slide

Derivatives

Question

$If f (x) = \log (\frac{1 + x}{1 - x}) and g (x) = \frac{3 x + x^{3}}{1 + 3 x^{2}} then \frac{d}{d x} (f (g (x))) =$

Moderate

A

$- f^{1} (x)$
B

$2 f^{1} (x)$
C

$3 f^{1} (x)$
D

$4 f^{1} (x)$

Solution

$\begin{array}{l} f (g (x)) = f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) \\ == \log (\frac{1 + \frac{3 x + x^{3}}{1 + 3 x^{2}}}{1 - \frac{3 x + x^{3}}{1 + 3 x^{2}}}) = \log {(\frac{1 + x}{1 - x})}^{3} = 3 \log (\frac{1 + x}{1 - x}) \\ = 3 f (x) ∴ \frac{d}{d x} (f (g (x))) = 3 f^{'} (x) \end{array}$

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