In Exercises 7–38, find the derivative of y with respect to x,...

Question

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.

33. y = ln(sec(lnθ))

Accepted Answer

Welcome back, everyone. Find D Y divided by DX4 Y equals LN of cosine of LN X. For this problem, we want to find the derivative of Y with respect to X, and we can notice that this is a composite function. So we're going to apply the chain rule. Let's remember that first of all, we are differentiating the outer function, which is LN of cosine of LN X. With respect to the first inner function. So that would be D of, cosine of LN of X, right? According to the chain rule, we are then multiplying by the derivative of the inner function, so that'd bed of cosine of L and X with respect to the next in function, so that would be LN X. So we're dividing by D of LN X. And finally, we're going to multiply by the derivative of L and of X with respect to X, right? Because X is the independent variable, we're going to stop here. So now let's evaluate each derivative. Let's remember that the derivative of LN of X is one divided by X. The main difference is that now instead of X, we have cosine of LN X. So that's how we get our first derivative. Now, for the second one, we're differentiating cosine. The derivative of cosine is negative sign, so we're going to get negative sign. Of L and of X. And finally, the derivative of LN of X is one divided by X. Now, let's go ahead and simplify. We're going to get negative sine divided by cosine, which is negative tangents of LN X, and we are multiplying by 1 divided by X. Overall, we can write it as negative 1 divided by X, multiplied by tangent of L and of X. That's our final answer, and thank you for watching.

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.33. y = ln(sec(lnθ))

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In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
33. y = ln(sec(lnθ))