Question Video: Differentiating Root Functions Using the Chain Rule | Nagwa Question Video: Differentiating Root Functions Using the Chain Rule | Nagwa

Question Video: Differentiating Root Functions Using the Chain Rule Mathematics • Second Year of Secondary School

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Determine the derivative of 𝑓(π‘₯) = 2√(2π‘₯ βˆ’ 1).

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Video Transcript

Determine the derivative of 𝑓 of π‘₯ is equal to two times the square root of two π‘₯ minus one.

The question wants us to find an expression for the derivative of our function 𝑓 of π‘₯. And we can see that 𝑓 of π‘₯ is two times the square root of two π‘₯ minus one. And the square root of two π‘₯ 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 π‘₯ which is the composition of two functions 𝑒 and 𝑣, then the chain rule tells us 𝑓 prime of π‘₯ is equal to 𝑣 prime of π‘₯ times 𝑒 prime evaluated at 𝑣 of π‘₯. So let’s write our function 𝑓 of π‘₯ in this form.

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

To use the chain rule, we need to find 𝑣 prime and 𝑒 prime. Let’s start by finding 𝑣 prime of π‘₯. That’s the derivative of two π‘₯ minus one with respect to π‘₯. And of course, this is a linear function, so its slope is equal to the coefficient of π‘₯, which is two. We could’ve also evaluated this by using the power rule for differentiation. Let’s now find an expression for 𝑒 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 𝑒 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 π‘₯ and 𝑒 prime of 𝑣. This means we’re now ready to use our chain rule to evaluate 𝑓 prime of π‘₯.

The chain rule tells us 𝑓 prime of π‘₯ is equal to 𝑣 prime of π‘₯ times 𝑒 prime evaluated at 𝑣 of π‘₯. We showed that 𝑣 prime of π‘₯ is equal to two. And we showed that 𝑒 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 π‘₯ minus one. This gives us that 𝑓 prime of π‘₯ is equal to two times one divided by the square root of two π‘₯ minus one, which we can simplify to be two divided by the square root of two π‘₯ minus one. Therefore, we’ve shown if 𝑓 of π‘₯ is equal to two times the square root of two π‘₯ minus one, then 𝑓 prime of π‘₯ is equal to two divided by the square root of two π‘₯ minus one.

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