# Question Video: Differentiating Combinations of Logarithmic and Polynomial Functions Using the Product Rule Mathematics • Higher Education

Find dπ¦/dπ₯ if π¦ = (6π₯βΉ β 5) ln π₯.

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### Video Transcript

Find dπ¦ by dπ₯ if π¦ is equal to six times π₯ to the ninth power minus five all multiplied by the natural logarithm of π₯.

Weβre given an equation for π¦ in terms of π₯. And we need to use this to find an expression for dπ¦ by dπ₯. Thatβs the derivative of π¦ with respect to π₯. To do this, we need to notice that π¦ is the product of two functions. Itβs six π₯ to the ninth power minus five multiplied by the natural logarithm of π₯. And we know how to differentiate both of these expressions. So we can evaluate this derivative by using the product rule.

So letβs start by recalling the product rule. We know if π¦ is the product of two differentiable functions π of π₯ times π of π₯, then the product rule tells us dπ¦ by dπ₯ would be equal to π prime of π₯ times π of π₯ plus π prime of π₯ times π of π₯. And we can see this is exactly what we have in this question. Weβll set π of π₯ to be six π₯ to the ninth power minus five and π of π₯ to be the natural logarithm of π₯.

So we can find dπ¦ by dπ₯ by using the product rule. However, we see to use the product rule, weβre going to need to find expressions for π prime of π₯ and π prime of π₯. Letβs start by finding an expression for π prime of π₯. Thatβs the derivative of π of π₯ with respect to π₯, which is the derivative of six π₯ to the ninth power minus five with respect to π₯. And we can see this is the derivative of a polynomial. So we can do this term by term by using the power rule for differentiation.

We want to multiply by our exponent of π₯ and reduce this exponent by one. In our first term, this gives us nine times six π₯ to the eighth power. And in our second term, we know the derivative of the constant negative five with respect to π₯ will be equal to zero. And of course, we can simplify this expression for π prime of π₯. We get nine times six is equal to 54. So weβve shown π prime of π₯ is equal to 54π₯ to the eighth power.

Letβs now find an expression for π prime of π₯. Once again, thatβs the derivative of π of π₯ with respect to π₯, which is the derivative of the natural logarithm of π₯ with respect to π₯. And to evaluate this derivative, we need to recall one of our standard derivative results. The derivative of the natural logarithm of π₯ with respect to π₯ is equal to the reciprocal function one over π₯. So by using this, weβve shown that π prime of π₯ is equal to one over π₯.

Weβre now ready to find an expression for dπ¦ by dπ₯ by using our formula from the product rule. We just need to substitute our expressions for π of π₯, π of π₯, π prime of π₯, and π prime of π₯ into our formula from the product rule. Doing this, we get dπ¦ by dπ₯ is equal to 54π₯ to the eighth power times the natural logarithm of π₯ plus one over π₯ multiplied by six π₯ to the ninth power minus five.

And we could leave our answer like this. However, we could also simplify. Weβll do this by distributing one over π₯ over the parentheses in our second term. This gives us 54π₯ to the eighth power multiplied by the natural logarithm of π₯ plus six π₯ to the ninth power divided by π₯ minus five over π₯. And of course we can simplify this. We have six π₯ to the ninth power divided by π₯. Using our laws of exponents, this is just equal to six π₯ to the eighth power.

And thereβs one more piece of simplification we can do. We can take out the shared factor of six π₯ to the eighth power in our first and second term. And doing this gives us six π₯ to the eighth power multiplied by nine times the natural logarithm of π₯ plus one minus five over π₯. And this gives us our final answer.

Therefore, we were able to show if π¦ is equal to six π₯ to the ninth power minus five all multiplied by the natural logarithm of π₯, then by using the product rule, we can show that dπ¦ by dπ₯ must be equal to six π₯ to the eighth power multiplied by nine times the natural logarithm of π₯ plus one minus five divided by π₯.