Video: Differentiating Inverse Trigonometric Functions

Evaluate d/dπ‘₯ sin^{βˆ’1} (√(1 βˆ’ 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 we’ll use the chain rule to find its derivative. This says that if 𝑦 is some function in 𝑒 and 𝑒 is some function in π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. We’ll let 𝑒 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 𝑦 is equal to the inverse sin of 𝑒. To apply the chain rule, we’re going to need to find the derivative of both of these functions. The derivative of the inverse sin of 𝑒 with respect to 𝑒 is one over the square root of one minus 𝑒 squared.

And we can use the general power rule to find the derivative of one minus π‘₯ squared to the power of one-half. It’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.

d𝑦 by dπ‘₯ is, therefore, negative π‘₯ over the square root of one minus π‘₯ squared times one over the square root of one minus 𝑒 squared. We can replace 𝑒 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.

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