Question Video: The Derivative of an Inverse Sine Function | Nagwa Question Video: The Derivative of an Inverse Sine Function | Nagwa

Question Video: The Derivative of an Inverse Sine Function Mathematics • Higher Education

Find d/dπ‘₯ sin⁻¹ π‘₯.

04:22

Video Transcript

Find the derivative of the inverse sine function with respect to π‘₯.

We’re differentiating the inverse sin of π‘₯ with respect to π‘₯. So we’re going to begin by letting 𝑦 be equal to the inverse sin of π‘₯. Then we can say that π‘₯ must be equal to sin of 𝑦. We’re going to differentiate both sides of this equation with respect to π‘₯. So we say that d by dπ‘₯ of π‘₯ is equal to d by dπ‘₯ of sin of 𝑦. Well, the derivative of π‘₯ with respect to π‘₯ is quite straight forward; it’s one. But we’re going to need to use implicit differentiation which is a special case of the chain rule to differentiate sin 𝑦 with respect to π‘₯.

The derivative of sin 𝑦 with respect to 𝑦 is cos 𝑦. So the derivative of sin 𝑦 with respect to π‘₯ is cos 𝑦 times the derivative of 𝑦 with respect to π‘₯ which is just d𝑦 by dπ‘₯. So we currently see that one is equal to cos of 𝑦 times d𝑦 by dπ‘₯. We divide both sides of this equation by cos 𝑦 to form an equation for the derivative. And we see that d𝑦 by dπ‘₯ is equal to one over cos 𝑦. Now, we’ve got a bit of a problem. We do want an expression for the derivative in terms of π‘₯ not 𝑦.

And remember, we said that π‘₯ was equal to sin of 𝑦. So we’ll use the identity cos squared πœƒ plus sin squared πœƒ equals one. And I’ve replaced πœƒ with 𝑦. We’ll subtract sin squared 𝑦 from both sides of the equation. And then we’ll take the square root of both sides. And we see that cos of 𝑦 is equal to the positive and negative square root of one minus sin squared 𝑦. Remember, inverse sin is restricted to the closed interval negative πœ‹ by two to πœ‹ by two.

By our definition, that means 𝑦 must be greater than or equal to πœ‹ two and less than or equal to πœ‹ by two which, in turn, means cos of 𝑦 must be greater than or equal to zero and less than or equal to one. And that’s because in the interval 𝑦 is greater than or equal to negative πœ‹ by two and less than or equal to πœ‹ by two. The smallest value cos of 𝑦 we’ll take is zero. And the greatest value is one. And what this means here is we’re going to take the positive square root of one minus sin squared 𝑦 only.

We can now replace sin of 𝑦 with π‘₯. And we see that cos of 𝑦 is equal to the square root of one minus π‘₯ squared. And, therefore, d𝑦 by dπ‘₯ equals one over the square root of one minus π‘₯ squared. And we found the derivative of the inverse sin of π‘₯. It’s one over the square root of one minus π‘₯ squared for values of π‘₯ in the range π‘₯ is greater than negative one and less than one.

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