# Question Video: Differentiating Functions Involving Trigonometric Ratios Using Logarithmic Differentiation Mathematics

Use logarithmic differentiation to find the derivative of the function π¦ = 2(cos π₯)^π₯.

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### Video Transcript

Use logarithmic differentiation to find the derivative of the function π¦ equals two cos π₯ to the power of π₯.

We want to find the derivative of π¦ equals two cos π₯ to the power of π₯. In other words, we want to find ππ¦ by ππ₯. And weβre given a hint as to how to do this. Weβre told to use logarithmic differentiation. So what does logarithmic differentiation mean?

Well, the logarithmic part of a logarithmic differentiation is to take the natural logarithm on both sides of this equation. The differentiation part of logarithmic differentiation is to then differentiate both sides. But before we do that, letβs use some laws of logarithms to make the right-hand side easier to differentiate.

The first thing we can do is write the logarithm of a product as the sum of two logarithms. But this isnβt particularly helpful for the purposes of the logarithmic differentiation. The really helpful law of logarithms here is that the logarithm of π to the power of π is π times the logarithm of π. So with π equal to cos π₯ and π equal to π₯, we find out the natural logarithm of π¦ ln π¦ is ln two plus π₯ ln cos π₯.

Okay, now that weβve simplified as much as possible using the laws of logarithms, itβs time to move on to the differentiation part of logarithmic differentiation. And so we differentiate both sides with respect to π₯. Letβs clear some room to do this differentiating. How do we differentiate ln π¦ with respect to π₯? Well, we can apply the chain rule. With π equal to ln π¦, then the quantity weβre looking for is ππ by ππ₯.

And the chain rule tells us that this is ππ by ππ¦ times ππ¦ by ππ₯. So here, we just apply the chain rule with π equal to ln π¦. And what is π ln π¦ by ππ¦ β the derivative of ln π¦ with respect to π¦? Itβs just one over π¦. The derivative of the natural logarithm function is the reciprocal function.

Okay, now, we move on to the right-hand side, the derivative of a sum is the sum of the derivatives. And as the natural logarithm of two ln two is just a constant, its derivative with respect to π₯ is zero. We only have to worry about the other term. This is the derivative of the product of π₯ and ln cos π₯ with respect to π₯. So we should apply the product rule.

We apply the product rule. Take a moment here if youβd like to pause the video and check that weβve applied it correctly. And we can see that the second term that we get is relatively straightforward. π by ππ₯ of π₯ is just one. And so this term is just ln cos π₯. The other derivative is a bit more difficult. To compute it, weβre going to need the chain rule again.

If we let π equal ln cos π₯, then weβre looking for ππ by ππ₯. And letβs call our auxiliary variable in the chain rule π§. Set π§ equal to cos π₯ and then π is just ln π§. Here, Iβve just used the definition of π to write π by ππ₯ over ln cos π₯ as ππ by ππ₯. Now, we can apply the chain rule. ππ by ππ₯ is ππ by ππ§ times ππ§ by ππ₯. What is ππ by ππ§? As π is equal to ln π§, ππ by ππ§ is one over π§. And what was π§? Well, it was our auxiliary variable, which redefines to be cos π₯.

So letβs replace π§ by cos π₯ to get everything written in terms of π₯. Now, how about ππ§ by ππ₯? Well, π§ is still cos π₯ and the derivative of cos π₯ with respect to π₯ is minus sin π₯. So there we are, weβve applied the chain rule. We found that the derivative of ln cos π₯ is one over cos π₯ times minus sin π₯, which we can of course write as a single fraction. This is minus sin π₯ over cos π₯, which we can further simplify to minus 10π₯.

This is perhaps somewhat surprising and it turns out that this is a relatively useful fact later on. Anyway, remember weβre looking for ππ¦ by ππ₯. And we very nearly have it. On the left-hand side, we have one over π¦ ππ¦ by ππ₯. And so we can multiply both sides by π¦ to find ππ¦ by ππ₯ in terms of π₯ and π¦. Whenever possible, we like to give ππ¦ by ππ₯ in terms of π₯ alone. And it is possible here because we can substitute two times cos π₯ to the π₯ for π¦.

Making this substitution, rearranging somewhat, and writing ππ¦ by ππ₯ as π¦ prime, we get our final answer π¦ prime is two cos π₯ to the π₯ times ln cos π₯ minus π₯ 10π₯.

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