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Question

Mathematics Question on Continuity and differentiability

Differentiate the following w.r.t. x: esin−1xe^{sin^{-1}x}

Answer

let y = esin−1xe^{sin^{-1}x}
By using the chain rule, we obtain

dydx\frac {dy}{dx} = \frac {d}{dx}$$(e^{sin^{-1}x})

⇒$$\frac {dy}{dx} = esin−1xe^{sin^{-1}x} . ddx\frac {d}{dx}(sin-1x)

⇒$$\frac {dy}{dx} = esin−1xe^{sin^{-1}x} . 11−x2\frac {1}{\sqrt {1-x^2}}

⇒$$\frac {dy}{dx} = esin−1x1−x2\frac {e^{sin^{-1}x}}{\sqrt {1-x^2}}

∴dydx\frac {dy}{dx} = esin−1x1−x2\frac {e^{sin^{-1}x}}{\sqrt {1-x^2}}, x∈(-1,1)