# Video: Differentiating Root Functions Using the Chain Rule

If π¦ = the fifth root of (5π₯ + 2)β·, find dπ¦/dπ₯.

03:47

### Video Transcript

If π¦ is equal to the fifth root of five π₯ plus two to the seventh power, find dπ¦ by dπ₯.

The question wants us to find dπ¦ by dπ₯. Thatβs the first derivative of π¦ with respect to π₯. And we can see that π¦ is equal to the fifth root of the seventh power of a linear function. Thatβs the composition of three functions. So we could do this by using the chain rule twice. And this would work. However, thereβs a simpler method by using our laws of exponents. Weβll use the following two versions of our laws of exponents.

First, we know the fifth root of π is just equal to π to the power of one-fifth. Next, we know that π to the power of π all raised to the power of π is just equal to π to the power of π times π. Using these, we can show that π¦ is equal to five π₯ plus two raised to the power of seven over five. But now, we can see that π¦ is the composition of only two functions. Itβs a linear function raised to the power of seven over five. So we can differentiate this by using the chain rule. Weβll set our inner function five π₯ plus two to be equal to π’. This gives us that π¦ is equal to π’ to the power of seven over five.

We now recall the following version of the chain rule. If we have π¦ is a function of π’ and π’ is a function of π₯, then we can find the derivative of π¦ with respect to π₯ by first finding the derivative of π¦ with respect to π’ and multiplying this by the derivative of π’ with respect to π₯. And this is exactly what we have. We have that π¦ is a function of π’. Itβs π’ to the power of seven over five. And we have that π’ is a function of π₯. Itβs five π₯ plus two. So we need to find dπ¦ by dπ’ and dπ’ by dπ₯.

Letβs start by finding dπ¦ by dπ’. Thatβs the derivative of π’ to the power of seven over five with respect to π’. And we can do this by using the power rule for differentiation. We multiply by the exponent of π’, thatβs seven over five, and reduce this exponent by one. This gives us seven-fifths times π’ to the power of seven-fifths minus one. And seven-fifths minus one is equal to two-fifths. So we found an expression for dπ¦ by dπ’. Letβs now find an expression for dπ’ by dπ₯. We have that dπ’ by dπ₯ is the first derivative of π’ with respect to π₯.

And remember, π’ is equal to five π₯ plus two. So we want to differentiate five π₯ plus two with respect to π₯. And we can evaluate this by using the power rule for differentiation. Or we can notice that this is a linear function. So its slope is equal to the coefficient of π₯, which is five. So we found expressions for dπ¦ by dπ’ and dπ’ by dπ₯. We can now use the chain rule. The chain rule tells us dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. And we found that dπ¦ by dπ’ is equal to seven-fifths π’ to the power of two over five. And dπ’ by dπ₯ is equal to five. And we can simplify this by canceling the shared factor of five in the numerator and the denominator, giving us seven times π’ to the power of two over five.

But remember, weβre finding an expression for the derivative of π¦ with respect to π₯. So we want our answer to be in terms of π₯. So weβll use our substitution π’ is equal to five π₯ plus two. So by using the substitution π’ is equal to five π₯ plus two, weβve shown that dπ¦ by dπ₯ is equal to seven times five π₯ plus two to the power of two-fifths. And we could leave our answer like this. However, remember, our original expression for π¦ contained the fifth root of a linear function. So we can use both of our laws of exponents to write our answer in this form. Doing this, we get dπ¦ by dπ₯ is equal to seven times the fifth root of five π₯ plus two squared.

Therefore, by using the chain rule, weβve shown if π¦ is equal to the fifth root of five π₯ plus two to the seventh power, then dπ¦ by dπ₯ is equal to seven times the fifth root of five π₯ plus two squared.