WEBVTT
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Given that π¦ is equal to the log of eight π₯ raised to the power of four tan five π₯, find dπ¦ by dπ₯.
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Weβre asked to find the derivative of π¦ with respect to π₯, but our function π¦ is a logarithm which is raised to an exponent which is itself an expression in π₯.
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And because of this exponent, we canβt directly apply our usual tactics for differentiation, for example, the chain, product, or quotient rules.
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What we can do, however, is use whatβs called logarithmic differentiation.
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There are four steps to this method and weβre going to write these down first and then work through the steps for our function π¦.
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The first step is to apply the natural logarithms to both sides so that we have the natural logarithm of π¦ is the natural logarithm π of π₯, recalling that the natural logarithm is the logarithm to the base π, where π is Eulerβs number, and thatβs approximately 2.71828 and so on.
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We need also to specify that π¦ is greater than zero for this to be valid.
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Thatβs because the log of zero is undefined, and the logarithm doesnβt exist for negative values.
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If we do want to include negative values, then we need to use the absolute values for π¦ and π of π₯.
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And in this case, we specify that π¦ is nonzero.
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Our second step is to use the laws of logarithms to expand or simplify and this allows us to move on to step three.
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That is, we can differentiate both sides with respect to π₯.
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And our final step is to solve for dπ¦ by dπ₯.
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So now letβs apply this to our function π¦.
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For our first step, we have the natural logarithm of π¦ is the natural logarithm of the log of eight π₯ to the power four tan five π₯.
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And thatβs for π¦ greater than zero.
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So now we can move on to step two where we can use the laws of logarithms to expand our function.
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And since the argument of our natural logarithm involves an exponent, we can use the power rule for logarithms.
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This says that log to the base π of π raised to the power π is π times log to the base π of π.
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That is, we bring the exponent π down in front of the logarithm and multiply.
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Our exponent π is four tan five π₯.
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And so applying the power rule, we have four tan five π₯ times the natural logarithm of log of eight π₯.
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So far, so good because now we have a product of expressions on our right-hand side and we can move on to step three: that is, differentiate both sides with respect to π₯.
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On our right-hand side, we now have a product of two functions.
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This means we can use the product rule for differentiation.
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That is, for π’ and π£ differentiable functions of π₯, d by dπ₯ of the product π’π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.
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So now if we let π’ equal four tan five π₯ and π£ equal the natural logarithm of log eight π₯, we want to find dπ’ by dπ₯ and dπ£ by dπ₯ in order to use the product rule.
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Now for dπ’ by dπ₯, we can use the result that d by dπ₯ of tan π€, where π€ is a differentiable function of π₯, is dπ€ by dπ₯ times the sec squared of π€.
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Since we have π€ equal to five π₯, then dπ’ by dπ₯ is five times four sec squared five π₯.
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That is 26 sec squared five π₯.
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So now we want to find dπ£ by dπ₯.
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So we have a function π£ which is the natural logarithm of function of π₯, which weβll call π€.
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And π€ is the log of eight π₯.
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And remember that if we have a logarithm without the base specified, this means we have a log to the base 10.
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Now to differentiate our function, the natural logarithm of π€, we can use the result that d by dπ₯ of the natural logarithm of π€ is one over π€ dπ€ by dπ₯ where π€ is a function of π₯.
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And, of course, π€ is differentiable.
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So that we have dπ£ by dπ₯ is one over the log of eight π₯ times d by dπ₯ of the log of eight π₯.
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And now to differentiate the log to the base 10 of eight π₯, weβre going to use the rule for the derivative of the log to any base.
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This says that d by dπ₯ of log to the base π of π€ is equal to log to the base π of π all over π€ times dπ€ by dπ₯.
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So in our case, our base π is 10, and our function π€ is eight π₯.
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So its derivative is log to the base 10 of π over eight π₯, thatβs π€, times eight, which is dπ€ by dπ₯.
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The eights cancel each other out, and we have log to the base 10 of π over π₯.
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Okay, so this is fine.
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Now, can we simplify log to the base 10 of π?
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If we use the change of base formula for logarithms with π corresponding to π and 10 corresponding to π and so that our new base is π, we have log to the base 10 of π is equal to log to the base π of π over log to the base π of 10.
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And how is this simpler?
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Because log to the base π of π is one.
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And log to the base π of 10 is just the natural logarithm of 10.
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So log to the base 10 of π is actually one over the natural logarithm of 10 so that we have the derivative with respect to π₯ of log to the base 10 of eight π₯ is actually one over π₯ times the natural logarithm of 10.
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Then dπ£ by dπ₯ is then one over log eight π₯ times one over π₯ times the natural logarithm of 10.
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So now we have everything we need to use the product rule for differentiation.
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And if we make some room, we have π’ is four tan five π₯, π£ is the natural logarithm of log eight π₯ dπ’ by dπ₯ is 26 squared five π₯, and dπ£ by dπ₯ is one over π₯ times the natural logarithm of 10 times the logarithm of eight π₯.
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And into our product rule formula, we have d by dπ₯ of the natural logarithm of π¦ is four tan five π₯, which is π’, times one over π₯ times the natural logarithm of 10 times the log of eight π₯, which is dπ£ by dπ₯, plus the natural logarithm of the log of eight π₯, which is π£, times 20 sec squared five π₯, which is dπ’ by dπ₯.
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And rewriting this with the nonquotient first and rearranging, we have d by dπ₯ of the natural logarithm of π¦ is 20 times the natural logarithm of the log of eight π₯ times sec squared five π₯ plus four tan five π₯ over π₯ times the natural logarithm of 10 times log of eight π₯.
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And weβre almost finished but not quite since we still have to differentiate the left-hand side.
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We want d by dπ₯ of the natural logarithm of π¦.
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And remember that π¦ is a function of π₯, so we can again use the result that d by dπ₯ of the natural logarithm of π€, where π€ is a differentiable function of π₯, is one over π€ times dπ€ by dπ₯ for π€ greater than zero.
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And this gives us one over π¦ times dπ¦ by dπ₯.
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So weβve completed our step three; that is, weβve differentiated both sides with respect to π₯.
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So now we need to solve for dπ¦ by dπ₯.
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Thatβs step four.
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And we can do this by multiplying both sides by π¦.
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On our left-hand side, the π¦βs cancel to one, and on our right-hand side, we introduce our original function π¦.
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And weβve completed our step four.
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So for the function π¦ is log eight π₯ raised to the power four tan five π₯, dπ¦ by dπ₯ is log eight π₯ raised to the power four tan five π₯ times 20 times the natural logarithm of log eight π₯ times sec squared five π₯ plus four tan five π₯ over π₯ times the natural logarithm of 10 times log eight π₯.