Video: Differentiating Root Functions Using the Chain Rule

If 𝑦 = the fifth root of (5π‘₯ + 2)⁷, find d𝑦/dπ‘₯.

03:47

Video Transcript

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

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

First, we know the fifth root of π‘Ž is just equal to π‘Ž to the power of one-fifth. Next, we know that π‘Ž to the power of 𝑏 all raised to the power of 𝑐 is just equal to π‘Ž to the power of 𝑏 times 𝑐. Using these, we can show that 𝑦 is equal to five π‘₯ plus two raised to the power of seven over five. But now, we can see that 𝑦 is the composition of only two functions. It’s a linear function raised to the power of seven over five. So we can differentiate this by using the chain rule. We’ll set our inner function five π‘₯ plus two to be equal to 𝑒. This gives us that 𝑦 is equal to 𝑒 to the power of seven over five.

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

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

And remember, 𝑒 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 d𝑦 by d𝑒 and d𝑒 by dπ‘₯. We can now use the chain rule. The chain rule tells us d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. And we found that d𝑦 by d𝑒 is equal to seven-fifths 𝑒 to the power of two over five. And d𝑒 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 𝑒 to the power of two over five.

But remember, we’re finding an expression for the derivative of 𝑦 with respect to π‘₯. So we want our answer to be in terms of π‘₯. So we’ll use our substitution 𝑒 is equal to five π‘₯ plus two. So by using the substitution 𝑒 is equal to five π‘₯ plus two, we’ve shown that d𝑦 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 𝑦 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 d𝑦 by dπ‘₯ is equal to seven times the fifth root of five π‘₯ plus two squared.

Therefore, by using the chain rule, we’ve shown if 𝑦 is equal to the fifth root of five π‘₯ plus two to the seventh power, then d𝑦 by dπ‘₯ is equal to seven times the fifth root of five π‘₯ plus two squared.

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