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Use logarithmic differentiation to find the derivative of the function π¦ equals two cos π₯ to the power of π₯.
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We want to find the derivative of π¦ equals two cos π₯ to the power of π₯.
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In other words, we want to find ππ¦ by ππ₯.
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And weβre given a hint as to how to do this.
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Weβre told to use logarithmic differentiation.
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So what does logarithmic differentiation mean?
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Well, the logarithmic part of a logarithmic differentiation is to take the natural logarithm on both sides of this equation.
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The differentiation part of logarithmic differentiation is to then differentiate both sides.
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But before we do that, letβs use some laws of logarithms to make the right-hand side easier to differentiate.
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The first thing we can do is write the logarithm of a product as the sum of two logarithms.
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But this isnβt particularly helpful for the purposes of the logarithmic differentiation.
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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.
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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₯.
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Okay, now that weβve simplified as much as possible using the laws of logarithms, itβs time to move on to the differentiation part of logarithmic differentiation.
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And so we differentiate both sides with respect to π₯.
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Letβs clear some room to do this differentiating.
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How do we differentiate ln π¦ with respect to π₯?
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Well, we can apply the chain rule.
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With π equal to ln π¦, then the quantity weβre looking for is ππ by ππ₯.
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And the chain rule tells us that this is ππ by ππ¦ times ππ¦ by ππ₯.
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So here, we just apply the chain rule with π equal to ln π¦.
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And what is π ln π¦ by ππ¦ β the derivative of ln π¦ with respect to π¦?
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Itβs just one over π¦.
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The derivative of the natural logarithm function is the reciprocal function.
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Okay, now, we move on to the right-hand side, the derivative of a sum is the sum of the derivatives.
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And as the natural logarithm of two ln two is just a constant, its derivative with respect to π₯ is zero.
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We only have to worry about the other term.
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This is the derivative of the product of π₯ and ln cos π₯ with respect to π₯.
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So we should apply the product rule.
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We apply the product rule.
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Take a moment here if youβd like to pause the video and check that weβve applied it correctly.
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And we can see that the second term that we get is relatively straightforward.
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π by ππ₯ of π₯ is just one.
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And so this term is just ln cos π₯.
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The other derivative is a bit more difficult.
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To compute it, weβre going to need the chain rule again.
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If we let π equal ln cos π₯, then weβre looking for ππ by ππ₯.
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And letβs call our auxiliary variable in the chain rule π§.
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Set π§ equal to cos π₯ and then π is just ln π§.
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Here, Iβve just used the definition of π to write π by ππ₯ over ln cos π₯ as ππ by ππ₯.
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Now, we can apply the chain rule.
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ππ by ππ₯ is ππ by ππ§ times ππ§ by ππ₯.
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What is ππ by ππ§?
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As π is equal to ln π§, ππ by ππ§ is one over π§.
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And what was π§?
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Well, it was our auxiliary variable, which redefines to be cos π₯.
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So letβs replace π§ by cos π₯ to get everything written in terms of π₯.
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Now, how about ππ§ by ππ₯?
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Well, π§ is still cos π₯ and the derivative of cos π₯ with respect to π₯ is minus sin π₯.
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So there we are, weβve applied the chain rule.
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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.
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This is minus sin π₯ over cos π₯, which we can further simplify to minus 10π₯.
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This is perhaps somewhat surprising and it turns out that this is a relatively useful fact later on.
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Anyway, remember weβre looking for ππ¦ by ππ₯.
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And we very nearly have it.
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On the left-hand side, we have one over π¦ ππ¦ by ππ₯.
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And so we can multiply both sides by π¦ to find ππ¦ by ππ₯ in terms of π₯ and π¦.
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Whenever possible, we like to give ππ¦ by ππ₯ in terms of π₯ alone.
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And it is possible here because we can substitute two times cos π₯ to the π₯ for π¦.
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Making this substitution, rearranging somewhat, and writing ππ¦ by ππ₯ as π¦ prime, we get our final answer π¦ prime is two cos π₯ to the π₯ times ln cos π₯ minus π₯ 10π₯.