Video: The Derivative of an Inverse Cosine Function

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

02:28

Video Transcript

Find the derivative of the inverse cos of π‘₯ over π‘Ž with respect to π‘₯, where π‘Ž is not equal to zero.

We’ll begin by letting 𝑦 be equal to the inverse cos of π‘₯ over π‘Ž. This can be alternately written as π‘₯ over π‘Ž equals cos of 𝑦. And we can then multiply both sides by π‘Ž. And we see that π‘₯ is equal to π‘Ž times cos of 𝑦. We’re going to differentiate our expression for π‘₯ with respect to 𝑦. In other words, we’re going to find dπ‘₯ by d𝑦. We’ll use the general result that the derivative of cos of π‘₯ with respect to π‘₯ is negative sin of π‘₯. And we see that dπ‘₯ by d𝑦 must be equal to negative π‘Ž sin of 𝑦.

Now, before we perform the next step we need to recall the fact that for the inverse trigonometric functions, we restrict their domains. And we know that the domain of the inverse cos of π‘₯ or our cos of π‘₯ is greater than or equal to zero and less than or equal to πœ‹. This means that 𝑦 must be greater than or equal to zero and less than or equal to πœ‹. Now, we’re going to use the inverse function theorem. So we’re going to take values of 𝑦 greater than zero and less than πœ‹, such that sin of 𝑦 is not equal to zero.

Using this criteria, we can use dπ‘₯ by d𝑦 equals one over d𝑦 by dπ‘₯ which can be rearranged to say that d𝑦 by dπ‘₯ equals one over dπ‘₯ by d𝑦. And we see that, for our case, d𝑦 by dπ‘₯ equals one over negative π‘Ž sin of 𝑦. And we have an expression for the derivative in terms of 𝑦. Remember, we want this to be in terms of π‘₯. We said that π‘₯ over π‘Ž equals cos of 𝑦. So we’ll use the fact that sin squared 𝑦 plus cos squared 𝑦 equals one and rearrange this to say that sin 𝑦 equals plus or minus the square root of one minus cos squared 𝑦. When 𝑦 is between zero and πœ‹, sin 𝑦 is greater than zero. So in fact, we’re only interested in the positive root.

So we’ll replace this in our expression for the derivative. And we get negative one over π‘Ž times the square root of one minus cos squared 𝑦. We then replace cos 𝑦 with π‘₯ over π‘Ž and change π‘₯ over π‘Ž all squared to π‘₯ squared over π‘Ž squared. And then we bring π‘Ž into the square root. And we see that d𝑦 by dπ‘₯ is equal to negative one over the square root of π‘Ž squared minus π‘₯ squared. So d by dπ‘₯ of the inverse cos of π‘₯ over π‘Ž is equal to negative one over the square root of π‘Ž squared minus π‘₯ squared for values of π‘₯ between negative π‘Ž and π‘Ž.

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