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 π(π₯) = 2β(2π₯ β 1).

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### Video Transcript

Determine the derivative of π 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 π of π₯. And we can see that π 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. Weβre taking the square root of a linear function. So weβ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 π of π₯ which is the composition of two functions π’ and π£, then the chain rule tells us π prime of π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯. So letβs write our function π of π₯ in this form.

Our innermost function is the linear function two π₯ minus one. So weβll set π£ of π₯ to be equal to two π₯ minus one. Using this, we have π of π₯ is equal to two times the square root of π£ of π₯. We see weβre taking the square root of π£ of π₯ and then multiplying this by two. So to write π of π₯ as the composition of π’ and π£, weβll need π’ evaluated at π£ to be equal to two times the square root of π£. In other words, for these functions, π’ and π£, we have that π of π₯ is equal to π’ evaluated at π£ 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 π prime of π₯.

To use the chain rule, we need to find π£ prime and π’ prime. Letβs start by finding π£ prime of π₯. Thatβ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 couldβve also evaluated this by using the power rule for differentiation. Letβs now find an expression for π’ prime of π£. Thatβs the derivative of two times the square root of π£ with respect to π£. We can do this by using our laws of exponents. We know that two root π£ is equal to two times π£ to the power of one-half.

We can then differentiate this by using the power rule for differentiation. We multiply by the exponent of π£, thatβs one-half, and then reduce this exponent by one. This gives us one-half times two times π£ 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 π’ prime of π£ is equal to π£ to the power of negative one-half. Using our laws of exponents, we can write this as one divided by the square root of π£. So weβve now found expressions for π£ prime of π₯ and π’ prime of π£. This means weβre now ready to use our chain rule to evaluate π prime of π₯.

The chain rule tells us π prime of π₯ is equal to π£ prime of π₯ times π’ prime evaluated at π£ of π₯. We showed that π£ prime of π₯ is equal to two. And we showed that π’ prime of π£ is equal to one divided by the square root of π£. Finally, we want to use our substitution π£ is equal to two π₯ minus one. This gives us that π 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, weβve shown if π of π₯ is equal to two times the square root of two π₯ minus one, then π prime of π₯ is equal to two divided by the square root of two π₯ minus one.

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