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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Evaluate 𝑓′(βˆ’1), where 𝑓(π‘₯) = √(βˆ’8π‘₯Β² + 13).

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Video Transcript

Evaluate 𝑓 prime of negative one, where 𝑓 of π‘₯ is equal to the square root of negative eight π‘₯ squared plus 13.

We need to evaluate 𝑓 prime of negative one. Remember, since 𝑓 is a function in π‘₯, this means the derivative of 𝑓 with respect to π‘₯ evaluated when π‘₯ is equal to negative one. So we’re going to need to differentiate 𝑓 of π‘₯ with respect to π‘₯. And we can see that our function 𝑓 of π‘₯ is the composition of two functions. We’re taking the square root of a quadratic polynomial. And because this is the composition of two functions, we might want to use the chain rule, and this would work.

However, if we first simplify our function 𝑓 of π‘₯ by using our laws of exponents β€” specifically, the square root of π‘Ž is equal to π‘Ž to the power of one-half β€” then we can write 𝑓 of π‘₯ as negative eight π‘₯ squared plus 13 all raised to the power of one-half. And now, we can differentiate this by using the general power rule. Let’s start by recalling what the general power rule tells us. The general power rule is a specific version of the chain rule. It tells us for any real constant 𝑛 and differentiable function 𝑔 of π‘₯, the derivative of 𝑔 of π‘₯ all raised to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ all raised to the power of 𝑛 minus one.

In this case, we can see our function 𝑔 of π‘₯ will be negative eight π‘₯ squared plus 13 which we know is differentiable because it’s a polynomial. And our value of 𝑛 will be equal to one-half. So by applying the general power rule with 𝑛 equal to one-half and 𝑔 of π‘₯ equal to negative eight π‘₯ squared plus 13, we get that 𝑓 prime of π‘₯ is equal to one-half multiplied by 𝑔 prime of π‘₯ times negative eight π‘₯ squared plus 13 all raised to the power of one-half minus one. And we could start simplifying this. However, let’s first find an expression for 𝑔 prime of π‘₯.

𝑔 prime of π‘₯ will be the derivative of negative eight π‘₯ squared plus 13 with respect to π‘₯. And because this is a polynomial, we can do this term by term by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and then reduce this exponent by one. If we do this, we get negative 16π‘₯. Now that we’ve found this, let’s clear some space and use this to simplify our expression for 𝑓 prime of π‘₯. Substituting 𝑔 prime of π‘₯ is equal to negative 16π‘₯ and simplifying our exponent of one-half minus one to be equal to negative one-half, we get 𝑓 prime of π‘₯ is equal to one-half times negative 16π‘₯ multiplied by negative eight π‘₯ squared plus 13 all raised to the power of negative one-half.

And we can simplify this further. First, one-half multiplied by negative 16π‘₯ is equal to negative eight. Next, we’ll simplify this further by using our laws of exponents. Remember, raising a number to the power of negative one-half is the same as dividing by the square root of that number. So by using this, we’ve simplified 𝑓 prime of π‘₯ to be equal to negative eight π‘₯ divided by the square root of negative eight π‘₯ squared plus 13. But remember, the question didn’t want us to just find an expression for 𝑓 prime of π‘₯. We need to find 𝑓 prime evaluated at π‘₯ is equal to negative one. So we need to substitute π‘₯ is equal to negative one into this expression for 𝑓 prime of π‘₯.

Substituting π‘₯ is equal to negative one into our expression for 𝑓 prime of π‘₯, we get 𝑓 prime of negative one is equal to negative eight times negative one divided by the square root of negative eight times negative one squared plus 13. And if we evaluate this, we get eight divided by the square root of five. And we could leave our answer like this; however, we can also simplify by rationalizing our denominator. We’ll multiply both our numerator and our denominator by the square root of five. And by doing this, we get eight times the square root of five all divided by five.

Therefore, we were able to show if 𝑓 of π‘₯ is equal to the square root of negative eight π‘₯ squared plus 13, then, by using the general power rule, we were able to show that 𝑓 prime evaluated at negative one will be equal to eight times the square root of five all divided by five.

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