Question Video: Logarithmic Differentiation of Functions Involving Logarithms and Trigonometric Ratios | Nagwa Question Video: Logarithmic Differentiation of Functions Involving Logarithms and Trigonometric Ratios | Nagwa

# Question Video: Logarithmic Differentiation of Functions Involving Logarithms and Trigonometric Ratios Mathematics • Third Year of Secondary School

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Given that π¦ = (log 8π₯)^(4 tan 5π₯), find dπ¦/dπ₯.

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### Video Transcript

Given that π¦ is equal to the log of eight π₯ raised to the power of four tan five π₯, find dπ¦ by dπ₯.

Weβre 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₯. And because of this exponent, we canβt directly apply our usual tactics for differentiation, for example, the chain, product, or quotient rules. What we can do, however, is use whatβs called logarithmic differentiation. There are four steps to this method and weβre going to write these down first and then work through the steps for our function π¦.

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 Eulerβs number, and thatβs approximately 2.71828 and so on. We need also to specify that π¦ is greater than zero for this to be valid. Thatβs because the log of zero is undefined, and the logarithm doesnβt exist for negative values. If we do want to include negative values, then we need to use the absolute values for π¦ and π of π₯. And in this case, we specify that π¦ is nonzero.

Our second step is to use the laws of logarithms to expand or simplify and this allows us to move on to step three. That is, we can differentiate both sides with respect to π₯. And our final step is to solve for dπ¦ by dπ₯. So now letβs apply this to our function π¦. 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₯. And thatβs for π¦ greater than zero. So now we can move on to step two where we can use the laws of logarithms to expand our function. And since the argument of our natural logarithm involves an exponent, we can use the power rule for logarithms. 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 logarithm and multiply.

Our exponent π is four tan five π₯. And so applying the power rule, we have four tan five π₯ times the natural logarithm of log of eight π₯. 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₯. On our right-hand side, we now have a product of two functions. This means we can use the product rule for differentiation. 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₯. 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.

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€. Since we have π€ equal to five π₯, then dπ’ by dπ₯ is five times four sec squared five π₯. That is 26 sec squared five π₯. So now we want to find dπ£ by dπ₯. So we have a function π£ which is the natural logarithm of function of π₯, which weβll call π€. And π€ is the log of eight π₯. And remember that if we have a logarithm without the base specified, this means we have a log to the base 10.

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₯. And, of course, π€ is differentiable. 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₯. And now to differentiate the log to the base 10 of eight π₯, weβre going to use the rule for the derivative of the log to any base. 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₯. So in our case, our base π is 10, and our function π€ is eight π₯. So its derivative is log to the base 10 of π over eight π₯, thatβs π€, times eight, which is dπ€ by dπ₯. The eights cancel each other out, and we have log to the base 10 of π over π₯.

Okay, so this is fine. Now, can we simplify log to the base 10 of π? 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. And how is this simpler? Because log to the base π of π is one. And log to the base π of 10 is just the natural logarithm of 10. 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. Then dπ£ by dπ₯ is then one over log eight π₯ times one over π₯ times the natural logarithm of 10.

So now we have everything we need to use the product rule for differentiation. 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₯. 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₯.

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₯. And weβre almost finished but not quite since we still have to differentiate the left-hand side. We want d by dπ₯ of the natural logarithm of π¦. 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. And this gives us one over π¦ times dπ¦ by dπ₯.

So weβve completed our step three; that is, weβve differentiated both sides with respect to π₯. So now we need to solve for dπ¦ by dπ₯. Thatβs step four. And we can do this by multiplying both sides by π¦. On our left-hand side, the π¦βs cancel to one, and on our right-hand side, we introduce our original function π¦. And weβve completed our step four. 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₯.

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