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

Find d𝑦/dπ‘₯ if 𝑦 = (6π‘₯⁹ βˆ’ 5) ln π‘₯.

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Video Transcript

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

We’re given an equation for 𝑦 in terms of π‘₯. And we need to use this to find an expression for d𝑦 by dπ‘₯. That’s the derivative of 𝑦 with respect to π‘₯. To do this, we need to notice that 𝑦 is the product of two functions. It’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 let’s start by recalling the product rule. We know if 𝑦 is the product of two differentiable functions 𝑓 of π‘₯ times 𝑔 of π‘₯, then the product rule tells us d𝑦 by dπ‘₯ would be equal to 𝑓 prime of π‘₯ times 𝑔 of π‘₯ plus 𝑔 prime of π‘₯ times 𝑓 of π‘₯. And we can see this is exactly what we have in this question. We’ll set 𝑓 of π‘₯ to be six π‘₯ to the ninth power minus five and 𝑔 of π‘₯ to be the natural logarithm of π‘₯.

So we can find d𝑦 by dπ‘₯ by using the product rule. However, we see to use the product rule, we’re going to need to find expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Let’s start by finding an expression for 𝑓 prime of π‘₯. That’s the derivative of 𝑓 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 𝑓 prime of π‘₯. We get nine times six is equal to 54. So we’ve shown 𝑓 prime of π‘₯ is equal to 54π‘₯ to the eighth power.

Let’s now find an expression for 𝑔 prime of π‘₯. Once again, that’s the derivative of 𝑔 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, we’ve shown that 𝑔 prime of π‘₯ is equal to one over π‘₯.

We’re now ready to find an expression for d𝑦 by dπ‘₯ by using our formula from the product rule. We just need to substitute our expressions for 𝑓 of π‘₯, 𝑔 of π‘₯, 𝑓 prime of π‘₯, and 𝑔 prime of π‘₯ into our formula from the product rule. Doing this, we get d𝑦 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. We’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 there’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 𝑦 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 d𝑦 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‘₯.

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