# Video: Differentiating Inverse Trigonometric Functions

Evaluate d/dπ₯ sin^{β1} (β(1 β π₯Β²)).

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### Video Transcript

Evaluate the derivative of the inverse sin of the square root of one minus π₯ squared with respect to π₯.

Here, we have a function of a function or a composite function. So weβll use the chain rule to find its derivative. This says that if π¦ is some function in π’ and π’ is some function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. Weβll let π’ be equal to the square root of one minus π₯ squared. Which can, of course, alternatively be written as one minus π₯ squared to the power of one-half. Then π¦ is equal to the inverse sin of π’. To apply the chain rule, weβre going to need to find the derivative of both of these functions. The derivative of the inverse sin of π’ with respect to π’ is one over the square root of one minus π’ squared.

And we can use the general power rule to find the derivative of one minus π₯ squared to the power of one-half. Itβs a half times one minus π₯ squared to the negative one-half times the derivative of the bit inside the brackets which is negative two π₯. That can be written as negative π₯ times one minus π₯ squared to the power of negative one-half.

dπ¦ by dπ₯ is, therefore, negative π₯ over the square root of one minus π₯ squared times one over the square root of one minus π’ squared. We can replace π’ with one minus π₯ squared to the power of one-half. And the second fraction becomes one over the square root of one minus one minus π₯ squared. This further simplifies to one over π₯. And we divide through by π₯. And we see that the derivative of our function is negative one over the square root of one minus π₯ squared.