Video: The Derivative of an Inverse Sine Function

Find d/dπ‘₯ csc⁻¹ π‘₯.

02:16

Video Transcript

Find d by dπ‘₯ of the inverse cosecant of π‘₯.

We begin by letting 𝑦 be equal to the inverse cosecant of π‘₯. And this means we can rewrite this. And we can say that π‘₯ is equal to the cosecant of 𝑦.

We’re next going to use implicit differentiation to find the derivative of both sides of this equation. The derivative of π‘₯ with respect to π‘₯ is simply one. Then the derivative of cosec 𝑦 with respect to π‘₯ is equal to the derivative of cosec 𝑦 with respect to 𝑦 times d𝑦 by dπ‘₯. And the derivative of cosec 𝑦 with respect to 𝑦 is negative cosec 𝑦 cot 𝑦. So we see that one is equal to negative cosec 𝑦 cot 𝑦 times d𝑦 by dπ‘₯.

Now, we know that, for the inverse cosecant function, 𝑦 must be greater than negative πœ‹ by two and less than πœ‹ by two and not equal to zero. Using these restrictions cosec 𝑦 cot 𝑦 cannot be equal to zero. So we can divide through by negative cosec 𝑦 cot 𝑦. And we see that d𝑦 by dπ‘₯ is as shown. We want to represent our equation for the derivative in terms of π‘₯. So we’ll use this trigonometric identity cot squared 𝑦 plus one equals cosec squared 𝑦. And we can rewrite this to say that cot of 𝑦 is equal to the positive and negative square root of cosec squared 𝑦 minus one.

We put this into the equation for the derivative in place of cot of 𝑦. And we then use the fact that π‘₯ is equal to cosec 𝑦. But we are going to need to make a decision on the sine of the derivative. And it can help here to look at the graph of the inverse cosecant function. Notice how, for all values of π‘₯ in the range of the function, the derivative of the slope of the tangent is negative.

And we, therefore, use the absolute value to ensure that our derivative is always negative. We say that d𝑦 by dπ‘₯ is equal to the negative of the absolute value of one over π‘₯ times the square root of π‘₯ squared minus one. Since one and the square root of π‘₯ squared minus one are always positive, we can rewrite this as shown. So the derivative of the inverse cosecant function π‘₯ with respect to π‘₯ is negative one over the modulus or absolute value of π‘₯ times the square root of π‘₯ squared minus one.

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