Let f \left(\right. \theta \left.\right) = sin \left(\right. \left(tan\right)^{- 1} \left(\right. \frac{sinθ}{\sqrt{cos 2 \theta}} \left.\right) \left.\right) where \frac{- \pi}{4} < \theta < \frac{\pi}{4}. Then the value of \frac{d}{d \left(\right. tanθ \left.\right)} \left(\right. f \left(\right. \theta \left.\right) \left.\right), is :

Engineering

Mathematics

Derivative of Inverse Function

Question

Let $f (θ) = \sin (\tan^{- 1} (\frac{sinθ}{\sqrt{\cos 2 θ}})) where \frac{- π}{4} < θ < \frac{π}{4}$ . Then the value of $\frac{d}{d (tanθ)} (f (θ))$ , is :

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$∵ \tan (\frac{sinθ}{\sqrt{\cos 2 θ}}) = \sin^{- 1} tanθ$

so f ( $θ$ ) = sin (sin^–1 tan $θ$ ) = tan $θ$

$∵ \frac{d (f (θ))}{d (tanθ)} = \frac{d (tanθ)}{d (tanθ)} = 1$