Question

If $y = \sin^{-1}(x)$, then $y' = \frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1 - x^2}}$. This problem will walk you through the steps of calculating the derivative. (a) Use the definition of inverse to rewrite the given equation with $x$ as a function of $y$.

          If $y = \sin^{-1}(x)$, then $y' = \frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1 - x^2}}$.
This problem will walk you through the steps of calculating the derivative.
(a) Use the definition of inverse to rewrite the given equation with $x$ as a function of $y$.

If y = sin^-1(x), then y' = (d)/(dx)[sin^-1(x)] = (1)/(√(1 - x^2)).
This problem will walk you through the steps of calculating the derivative.
(a) Use the definition of inverse to rewrite the given equation with x as a function of y.

Added by Mark H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ma. Theresa Alin

04/29/2022

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y' = (-1)c sin y Show more…

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If y = sin-1(), then y' This problem will walk you through the steps of calculating the derivative. (a) Use the definition of inverse to rewrite the given equation with c as a function of y