In Exercises 1–24, find the derivative of y with respect to th...

Question

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.

23. y = arccsc(secθ), 0<θ<π/2

Accepted Answer

Welcome back everyone. If y equals inverse of sine of cosine of t for t between 0 and pi divided by 2 not inclusive, find the Dy divided by dt. So for this problem we can actually use a shortcut. Let's remember the identity cosine of t is equal to sine of pi divided by 2 minus c. So if we consider our function y is equal to inverse of sin of sin of pi divided by 2 minus t, and we can use the properties of inverse functions, right. Now let's set u equals pi divided by 2 minus c to simplify the expression just to see clearly what we have. So y can be written as inverse of sin of sin of u, right? We know that you. In the principal value, property belongs to the interval from gpi divided by 2 up to pi divided by 2, right? And we know that why. is going to be equal to pi divided by 2 minus c, right, because inverse of sin of sin of u is equal to u, so y equals u and therefore y equals pi divided by 2 minus c. And that's because pi divided by 2 minus t belongs to the interval. From 0 up to 1 divided by 2, which is. In the principal value. Therefore We know that y is equal to pi divided by 2 minus t, and we can find the d y divided by dt by simply finding the derivative of pi divided by 2 minus t. We take the derivative of the constant pi divided by 2 and subtract the derivative of c. The derivative of a constant is 0 and the derivative oft is 1, so we get. -1. That's our final answer, and thank you for watching.

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.23. y = arccsc(secθ), 0<θ<π/2

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In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
23. y = arccsc(secθ), 0<θ<π/2