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

Determine dπ¦/dπ₯ for the function π¦: π¦ = (3π₯ + 8)^(β2 cos 8π₯).

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### Video Transcript

Determine dπ¦ by dπ₯ for the function π¦ such that π¦ is equal to three π₯ plus eight all to the power negative two cos eight π₯.

Weβre asked to differentiate π¦ with respect to π₯. But notice that π¦ is an expression involving π₯ which is raised to the power of another expression which involves the variable π₯. And itβs because of this exponent, which is itself a function of π₯, that we canβt apply the usual tools of differentiation directly, such as the quotient, product, or chain rules. All is not lost, however, since we do have at least one method we can try which is logarithmic differentiation. There are four steps to this method.

And for a function π¦ is π of π₯, the first step is to apply the natural logarithm to both sides so that we have the natural logarithm of π¦ is the natural logarithm of π of π₯. For our function π¦ is three π₯ plus eight all to the power negative two cos eight π₯, this gives us the natural logarithm of π¦ is the natural logarithm of our expression in π₯ on the right-hand side. Now, at this point, we need to specify that π¦ is greater than zero. This is because the log of zero is undefined and the log doesnβt exist for negative values. If we want to include negative values, then we must use the absolute value signs around π¦ and π of π₯. And in that case, π¦ is nonzero. In our case, weβll simply specify that π¦ is greater than zero.

Now, at this point, you might be thinking, βBut hang on. This is more complicated than the original function.β And thatβs where our second step for logarithmic differentiation comes in. And that is that we use the laws of logarithms to expand or simplify. In our problem, we have a natural logarithm of a function containing an exponent. And to make this more amenable for differentiation, we can use the power rule for logarithms. This tells us 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.

And reminding ourselves that the natural logarithm has a base π, which is Eulerβs number, by the power rule on our right-hand side, we have negative two cos eight π₯, bringing the exponent down, times the natural logarithm of three π₯ plus eight. And now we start to see how we can solve our problem since on our right-hand side we have the product of two expressions in π₯ which we know how to differentiate. And this brings us to our third step in logarithmic differentiation, which is differentiate both sides with respect to π₯. And we can apply the product rule for differentiation on our right-hand side. And this tells us that if we have a product of two functions of π₯, π’ and π£, d by dπ₯ of π’ times π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.

And so we let π’ equal negative two cos eight π₯ and π£ equal the natural logarithm three π₯ plus eight. To differentiate π’, thatβs to find dπ’ by dπ₯, we can use the result that d by dπ₯ of cos π€, where π€ is a function of π₯, is negative dπ€ by dπ₯ sin π€, where in our case π€ is eight π₯. dπ€ by dπ₯ is eight. So we have negative eight times negative two sin eight π₯, that is, 16 sin eight π₯. And now to find dπ£ by dπ₯, we can use the result that d by dπ₯ of the natural logarithm of π€, where π€ is a function of π₯, is one of over π€ times dπ€ by dπ₯. And thatβs where π€ is greater than zero. And since π€ in this case is three π₯ plus eight, that means dπ€ by dπ₯ is three; dπ£ by dπ₯ is then one of the three π₯ plus eight times three, which is three over three π₯ plus eight.

So now we have everything we need to use the product rule for differentiation. And this gives us d by dπ₯ of the natural logarithm of π¦ on the left-hand side is negative two cos eight π₯, which is π’, times three over three π₯ plus eight, which is dπ£ by dπ₯, plus the natural logarithm of three π₯ plus eight, which is π£, times 16 sin eight π₯, which is dπ’ by dπ₯. And now making some room and rearranging so that our positive terms are at the front and combining the three and negative two in our second term, we have d by dπ₯ of the natural logarithm of π¦ is 16 sin eight π₯ times the natural logarithm of three π₯ plus eight minus six cos eight π₯ over three π₯ plus eight.

Weβre not quite finished, however, since we still need to differentiate our left-hand side. And again, we can use the known result for natural logarithm differentiation which says d by dπ₯ of natural logarithm of π€ is one over π€ dπ€ by dπ₯. Our function π¦ is actually a function of π₯, and so on our left-hand side we have one over π¦ dπ¦ by dπ₯. And now our final step, step four, is to solve for dπ¦ by dπ₯. We can do this by multiplying both sides by π¦ where, on our left-hand side, the π¦βs cancel each other out, leaving dπ¦ by dπ₯.

On our right-hand side, we reintroduce our original function π¦ so that for the function π¦ is three π₯ plus eight raised to the power negative two cos eight π₯, dπ¦ by dπ₯ is three π₯ plus eight raised to the power negative two cos eight π₯ times 16 sin eight π₯ times the natural logarithm of three π₯ plus eight minus six cos eight π₯ over three π₯ plus eight.