### Video Transcript

Find dπ¦ by dπ₯ if π¦ is equal to six π₯ raised to the power nine plus seven all raised to the power eight π₯.

Weβre asked to find the derivative with respect to π₯ of a compound function π¦ is equal to six π₯ raised to the power nine plus seven all raised to the power eight π₯. And if the exponent in our function was a constant, weβd be able to use the chain rule. However, our exponent is a function of π₯, and this means that none of the usual rules for differentiation apply. So weβre going to have to try something else to find dπ¦ by dπ₯. In fact, we can use a method called logarithmic differentiation.

How this works is if we have an expression π¦ is π of π₯, we take the natural logarithm on both sides. And remember that the natural logarithm is the logarithm to the base π where π is Eulerβs number which, to five decimal places, is 2.71828. And then using the fact that logs turn powers into product and product into sums, we use the laws of logarithms to expand. Our expression is then split into components which make it easier to differentiate. And once weβve differentiated, we can solve for dπ¦ by dπ₯. So letβs work through these steps for our expression.

Our first step is to take the natural logarithms on both sides. However, we do need to point out here that this is only valid for π¦ greater than zero. Remember that the logarithm of zero is undefined and the log function doesnβt exist for negative values. To cover negative values, we could take the natural logarithm of the absolute values of our expression. In this case, we can specify that π¦ is nonzero. However, in our case, weβll simply specify that π¦ is greater than zero. So weβve taken the natural logarithm on both sides, and our second step is to use the laws of logarithms to expand.

In our case, since our expression involves an exponent, we use the power rule for logarithms. This says that the logarithm of π raised to the power π is π times the logarithm of π. That is, we bring the exponent in front of our logarithm. Our exponent is eight π₯, so on our right-hand side, we have eight π₯ times the logarithm of six π₯ raised to the power nine plus seven. And although this looks quite complicated, in fact, we can now differentiate both sides. On the right-hand side, we can use the chain and the product rules. And on the left-hand side, we use implicit differentiation. Remember, we use implicit differentiation when weβre differentiating a function of π¦, not simply π¦ on its own. And in our case, we have the natural logarithm of π¦ which itself is a function of π₯.

So in general, if you have a function π which is a function of π¦ which is a function of π₯, we use the chain rule to find dπ by dπ₯ which is dπ by dπ¦ times dπ¦ by dπ₯. Weβre also going to use the result that d by dπ’ of the logarithm of π’ is one over π’ for π’ greater than zero. And using the notation π prime is dπ by dπ₯, the product rule for differentiation says the derivative of the product ππ is π prime π plus ππ prime. So on our left-hand side with π equal to the natural logarithm of π¦, and using our result for logarithms, we have the derivative of the natural logarithm of π¦ which is one over π¦ times dπ¦ by dπ₯.

So now we need to differentiate our right-hand side. On our right-hand side, if we let π’ equal to eight π₯, π£ equal to six π₯ raised to the power nine plus seven, and π€ equal to the natural logarithm of π£, then dπ’ by dπ₯ is equal to eight, dπ£ by dπ₯ is equal to 54 times π₯ raised to the power eight where here weβve used the power rule for differentiation. That is, if we have a function π times π₯ raised to the power π, then the derivative with respect to π₯ is πππ₯ raised to the π minus one. So we multiply by the exponent and subtract one from the exponent.

And to differentiate π€ with respect to π₯, we use the fact that π€ is a function of π£ which is a function of π₯, and therefore we use the chain rule. And since π€ is the natural logarithm of π£, weβre going to use our result for logarithms. So we have dπ€ by dπ£ is one over π£ and dπ£ by dπ₯ is 54π₯ raised to the power eight, which we just calculated. And since one over π£ is one over six π₯ raised to the power nine plus seven, dπ€ by dπ₯ is 54π₯ raised to the power eight over six π₯ raised to power nine plus seven. So now this is where our product rule comes in for our product π’ times π€.

So in our product rule, if we let π’ equal to π and π€ equal π, what we need to find is π’ prime π€ plus π’π€ prime. So just making some room. On our right-hand side, we want π’ prime times π€, which is eight times the natural logarithm of six π₯ raised to power nine plus seven, plus π’ times π€ prime, which is eight π₯ times 54π₯ raised to the power eight over six π₯ raised to the power nine plus seven. We have a common factor of eight, which we can take outside. And multiplying out our second term, our π₯-exponent raises to nine.

Now, remember, weβre trying to find dπ¦ by dπ₯. And our fourth step in logarithmic differentiation is to solve for dπ¦ by dπ₯. With one over π¦ dπ¦ by dπ₯ on the left-hand side, we can multiply through by π¦. And our π¦ on the left-hand side cancels. And so we have dπ¦ by dπ₯ equal to eight times π¦ multiplied by the natural logarithm of six π₯ raised to the power nine plus seven plus 54π₯ raised to the power nine over six π₯ raised to the power nine plus seven.