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.

17. y = ln(arccos(x))

Accepted Answer

Welcome back everyone. Differentiate y equals ln of inverse of cosine of 3 x. For this problem, we want to differentiate Y with respect to X, which is D Y divided by DX, right? We have a composite function, so let's apply the chain rule. First of all, we're going to take the outer function, which is LN. Of inverse of cosine of 3 x and we want to differentiate it with respect to the first inner function which is inverse of cosine of 3 x. So d of inverse of cosine of 3 x. Then according to the chain rule, we want to multiply by the derivative of the inner function. Inverse of cosine of 3 x. With respect to 3 X. And finally we are multiplying by the derivative of 3 x with respect to x. What we want to do is simply identify each derivative. Let's remember that the derivative of natural log is 1 divided by the inner function. So we're taking 1 divided by. Inverse of cosine of 3 x. That's our first derivative. Now for the second derivative, we want to differentiate inverse of cosine of 3 x with respect to 3 x. Let's remember that the derivative of inverse of cosine is -1 divided by a square root of 1. Minus the inner function squad, so we're going to take 3 x 2. And finally, the derivative of 3 x with respect to x is 3. We can factor out 3 and the derivative of x is 1, right? What we want to do now is simplify, so we're going to get negative. In the numerator we get 3, and in the denominator we're going to write inverse of cosine of 3 x multiplied by a square root of 1 minus 9 x squared. And that will be our final derivative for this problem. Let's label it and thank you for watching.

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.17. y = ln(arccos(x))

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In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
17. y = ln(arccos(x))