Question Video: Differentiating Functions Involving Trigonometric Ratios Using Logarithmic Differentiation | Nagwa Question Video: Differentiating Functions Involving Trigonometric Ratios Using Logarithmic Differentiation | Nagwa

Question Video: Differentiating Functions Involving Trigonometric Ratios Using Logarithmic Differentiation Mathematics • Third Year of Secondary School

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Determine dπ¦/dπ₯, given that π¦ = (8 sin 4π₯)^(2π₯).

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Video Transcript

Determine dπ¦ by dπ₯, given that π¦ is equal to eight times the sin of four π₯ all to the power two π₯.

Weβre asked to find the derivative of π¦ with respect to π₯. But if we look at our function π¦, we have an exponent two π₯, which is an expression in π₯. And this means that we canβt apply the usual rules of differentiation, the product, quotient, or chain rules, directly. What we can do, however, is use logarithmic differentiation. This is a four-step process for a function π¦ is π of π₯, where step one is to apply the natural logarithm to both sides, recalling that the natural logarithm is the log to the base π where π is Eulerβs number, which is approximately 2.71828 and so on.

In our case, we have π¦ is equal to eight times the sin of four π₯ all to the power two π₯, and so we take the natural logarithm on both sides. At this point, we are to specify that π¦ has to be greater than zero. Thatβs because the log of zero is undefined and the log doesnβt exist for negative values. If we do want to include negative values, we must include also the absolute value signs around π¦ and π of π₯ and, in this case, specify that π¦ is nonzero. For our problem, however, weβll simply specify that π¦ is greater than zero.

Now, our second step in logarithmic differentiation is where we start to see how taking logs can help us. We use the laws of logarithms to expand or simplify our right-hand side. And since in our argument for the logarithm on the right-hand side we have an exponent, weβre going to use the power rule for logarithms. And 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. That is, we bring the exponent π down in front of the log and multiply by it. In our case, this translates to the natural logarithm of π¦ is two π₯ times the natural logarithm of eight sin four π₯. And now we see that on our right-hand side, we have the product of two expressions in π₯.

And this brings us to our third step in logarithmic differentiation, thatβs differentiate both sides with respect to π₯. And since we have a product on our right-hand side, we can use the product rule for differentiation. This says that if π’ and π£ are differentiable functions of π₯, d by dπ₯ of π’π£, the product, is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. So if we now let π’ equal two π₯ and π£ be the natural algorithm of eight times the sin of four π₯, then to use the product rule, we need to find dπ£ by dπ₯ and dπ’ by dπ₯. dπ’ by dπ₯ is two by sight. And to differentiate π£, we can use the result that d by dπ₯ of the natural logarithm of a differentiable function π of π₯ is one over π of π₯ times dπ by dπ₯.

So dπ£ by dπ₯ is one over eight sin four π₯ times d by dπ₯ of eight sin four π₯. To differentiate eight sin four π₯, we use the fact that d by dπ₯ of sin π’, where π’ is a differentiable function of π₯, is dπ’ by dπ₯ times cos π’ so that dπ£ by dπ₯ is one over eight times the sin of four π₯ all times four times eight cos four π₯. We have a common factor of eight on the denominator and the numerator which cancel each other out so that dπ£ by dπ₯ is four cos four π₯ over sin four π₯. So we have π’ is two π₯ and dπ’ by dπ₯ is two, π£ is the natural algorithm of eight sin four π₯, and dπ£ by dπ₯ is four cos four π₯ over sin four π₯.

And to simplify this even further since cos over sin is the cot, we can say dπ£ by dπ₯ is four times the cot of four π₯. So now applying the product rule for differentiation to our right-hand side, we have two π₯, which is π’, times four times the cot of four π₯, which is dπ£ by dπ₯, plus the natural logarithm of eight sin four π₯, which is π£, times two, which is dπ’ by dπ₯. And now making some room, we can rewrite this and rearrange it. We have d by dπ₯ of the natural logarithm of π¦ is equal to two times the natural logarithm of eight sin four π₯ plus eight π₯ times the cot of four π₯.

Weβre not quite finished with our third step since we still need to differentiate the natural logarithm of π¦. And again, we can use our known result that d by dπ₯ of the natural logarithm of a function π of π₯ is one over π of π₯ times dπ by dπ₯. And since our π¦ is indeed a function of π₯, our left-hand side is simply one over π¦ dπ¦ by dπ₯. And this brings us to our final step in logarithmic differentiation which is to solve for dπ¦ by dπ₯.

We can do this by multiplying both sides by π¦ so that on the left-hand side, the π¦βs cancel each other out, and weβre left with dπ¦ by dπ₯. On our right-hand side, we can reintroduce our function π¦ so that weβre multiplying by eight sin four π₯ to the power two π₯ and taking out a common factor of two. Using logarithmic differentiation to differentiate the function π¦ is eight sin four π₯ all to the power two π₯, we have dπ¦ by dπ₯ is two times eight sin four π₯ to the power two π₯ multiplied by the natural logarithm of eight sin four π₯ plus four π₯ times the cot of four π₯.

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