First slide

Derivatives

Question

$\frac{d}{d x} \{\tan^{- 1} (\frac{a \sin x + b \cos x}{a \cos x - b \sin x})\} =$

Moderate

A

0
B

1
C

-1
D

$\frac{1}{1 + x^{2}}$

Solution

$p u t a = \cos α, b = \sin α, t h e n$
$\begin{array}{l} \tan^{- 1} (\frac{\sin (x + α)}{\cos (x + α)}) = \tan^{- 1} (\tan (x + α)) = x + α \\ ∴ \frac{d}{d x} (x + α) = 1 + 0 = 1 \end{array}$

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