First slide

Derivatives

Question

$If f (x) = \frac{x}{\sqrt{1 - x^{2}}}, g (x) = \frac{x}{\sqrt{1 + x^{2}}} then \frac{d}{d x} (f \circ g) (x) =$

Moderate

A

1
B

0
C

-1
D

2

Solution

$P u t x = \tan θ$
$(f o g) (x) = f (\frac{x}{\sqrt{1 + x^{2}}}) = f (\sin θ) = \frac{\sin θ}{\sqrt{1 - \sin^{2} θ}} = \tan θ = x \Rightarrow \frac{d}{d x} [(f o g) (x)] = 1$

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