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₯) sinβ»ΒΉ (π₯/π), where π β  0.

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### Video Transcript

Find d by dπ₯ of inverse sin of π₯ over π, where π is not equal to zero.

In this question, we need to differentiate the inverse or arcsin of π₯ over π with respect to π₯. We will begin this question by letting π¦ equal the inverse sin of π₯ over π. If we take the sin of both sides of this equation, we get sin π¦ is equal to π₯ over π. Our next step is to differentiate both sides of this equation with respect to π₯. We know that differentiating sin π₯ gives us cos π₯. This means that differentiating sin π¦ with respect to π₯ using our knowledge of implicit differentiation gives us cos π¦ multiplied by dπ¦ by dπ₯. Differentiating the right-hand side gives us one over the constant π. We can then divide both sides of this equation by cos π¦ such that dπ¦ by dπ₯ is equal to one over π multiplied by cos π¦.

We now have an expression for dπ¦ by dπ₯; however, this is not given in terms of π₯. If we go back to the equation sin π¦ is equal to π₯ over π, squaring both sides of this equation gives us sin squared π¦ is equal to π₯ squared over π squared. One of our trigonometric identities states that sin squared π plus cos squared π is equal to one. Rearranging this, we see that sin squared π is equal to one minus cos squared π. This means that sin squared π¦ is equal to one minus cos squared π¦. We now have one minus cos squared π¦ is equal to π₯ squared divided by π squared. Multiplying both sides by π squared gives us π squared minus π squared cos squared π¦ is equal to π₯ squared.

We can then rearrange this equation so that π squared minus π₯ squared is equal to π squared multiplied by cos squared π¦. Finally, we square root both sides of this equation. The right-hand side becomes π cos π¦. And the left-hand side is the square root of π squared minus π₯ squared. dπ¦ by dπ₯ is therefore equal to one over the square root of π squared minus π₯ squared. This is the derivative of inverse sin of π₯ over π.

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