# Question Video: Differentiating Root Functions Using the Chain Rule Mathematics • Higher Education

Determine the derivative of 𝑓(𝑥) = 2√(2𝑥 − 1).

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### Video Transcript

Determine the derivative of 𝑓 of 𝑥 is equal to two times the square root of two 𝑥 minus one.

The question wants us to find an expression for the derivative of our function 𝑓 of 𝑥. And we can see that 𝑓 of 𝑥 is two times the square root of two 𝑥 minus one. And the square root of two 𝑥 minus one is the composition of two functions. We’re taking the square root of a linear function. So we’re going to want to use the chain rule to evaluate this derivative. We recall the following version of the chain rule. If we have a function 𝑓 of 𝑥 which is the composition of two functions 𝑢 and 𝑣, then the chain rule tells us 𝑓 prime of 𝑥 is equal to 𝑣 prime of 𝑥 times 𝑢 prime evaluated at 𝑣 of 𝑥. So let’s write our function 𝑓 of 𝑥 in this form.

Our innermost function is the linear function two 𝑥 minus one. So we’ll set 𝑣 of 𝑥 to be equal to two 𝑥 minus one. Using this, we have 𝑓 of 𝑥 is equal to two times the square root of 𝑣 of 𝑥. We see we’re taking the square root of 𝑣 of 𝑥 and then multiplying this by two. So to write 𝑓 of 𝑥 as the composition of 𝑢 and 𝑣, we’ll need 𝑢 evaluated at 𝑣 to be equal to two times the square root of 𝑣. In other words, for these functions, 𝑢 and 𝑣, we have that 𝑓 of 𝑥 is equal to 𝑢 evaluated at 𝑣 evaluated at 𝑥, which is two times the square root of two 𝑥 minus one. So we can now apply the chain rule to find 𝑓 prime of 𝑥.

To use the chain rule, we need to find 𝑣 prime and 𝑢 prime. Let’s start by finding 𝑣 prime of 𝑥. That’s the derivative of two 𝑥 minus one with respect to 𝑥. And of course, this is a linear function, so its slope is equal to the coefficient of 𝑥, which is two. We could’ve also evaluated this by using the power rule for differentiation. Let’s now find an expression for 𝑢 prime of 𝑣. That’s the derivative of two times the square root of 𝑣 with respect to 𝑣. We can do this by using our laws of exponents. We know that two root 𝑣 is equal to two times 𝑣 to the power of one-half.

We can then differentiate this by using the power rule for differentiation. We multiply by the exponent of 𝑣, that’s one-half, and then reduce this exponent by one. This gives us one-half times two times 𝑣 to the power of one-half minus one. We can simplify this expression. One-half times two is equal to one, and one-half minus one is equal to negative one-half. So this gives us that 𝑢 prime of 𝑣 is equal to 𝑣 to the power of negative one-half. Using our laws of exponents, we can write this as one divided by the square root of 𝑣. So we’ve now found expressions for 𝑣 prime of 𝑥 and 𝑢 prime of 𝑣. This means we’re now ready to use our chain rule to evaluate 𝑓 prime of 𝑥.

The chain rule tells us 𝑓 prime of 𝑥 is equal to 𝑣 prime of 𝑥 times 𝑢 prime evaluated at 𝑣 of 𝑥. We showed that 𝑣 prime of 𝑥 is equal to two. And we showed that 𝑢 prime of 𝑣 is equal to one divided by the square root of 𝑣. Finally, we want to use our substitution 𝑣 is equal to two 𝑥 minus one. This gives us that 𝑓 prime of 𝑥 is equal to two times one divided by the square root of two 𝑥 minus one, which we can simplify to be two divided by the square root of two 𝑥 minus one. Therefore, we’ve shown if 𝑓 of 𝑥 is equal to two times the square root of two 𝑥 minus one, then 𝑓 prime of 𝑥 is equal to two divided by the square root of two 𝑥 minus one.