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

Find (d/dπ‘₯) sin⁻¹ (π‘₯/π‘Ž), where π‘Ž β‰  0.

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Video Transcript

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

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

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

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

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