Question Video: Differentiating Composite Functions Involving Roots given the Value of a Function and Its Derivative at a Point | Nagwa Question Video: Differentiating Composite Functions Involving Roots given the Value of a Function and Its Derivative at a Point | Nagwa

# Question Video: Differentiating Composite Functions Involving Roots given the Value of a Function and Its Derivative at a Point Mathematics • Second Year of Secondary School

## Join Nagwa Classes

Given that π¦ = β(π(π₯)), πβ²(4) = 2, and π(4) = 7, determine dπ¦/dπ₯ at π₯ = 4.

04:06

### Video Transcript

Given that π¦ is equal to the square root of π of π₯, π prime evaluated at four is equal to two, and π evaluated at four is equal to seven, determine dπ¦ by dπ₯ at π₯ is equal to four.

We need to determine the value of dπ¦ by dπ₯ when π₯ is equal to four. Weβre told that π¦ is equal to the square root of π of π₯. Remember, dπ¦ by dπ₯ will be the derivative of π¦ with respect to π₯. So, we need to find an expression for the derivative of the square root of π of π₯ with respect to π₯. One way of doing this is to notice that π¦ is given as the composition of two functions; weβre taking the square root of π of π₯. So, we could evaluate this derivative by using the chain rule, and this would work.

However, there is another method. We need to notice we can rewrite this expression by using our laws of exponents. We know the square root of π is equal to π to the power of one-half. So, we can rewrite our expression for π¦ as π of π₯ all raised to the power of one-half. Now, we have a function raised to the power of a constant, so we can evaluate the derivative of this by using the general power rule. We recall the general power rule tells us for any constant π and differentiable function π of π₯, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ all raised to the power of π minus one.

However, in this case, we might be worried about something. We want to set π of π₯ to be our inner function π of π₯. But to use the general power rule, we need π of π₯ to be differentiable and we donβt know that π is differentiable. However, we are told in the question that π prime of four is equal to two. So, we do know that π is differentiable when π₯ is equal to four. So, we can in fact apply the general power rule for π of π₯ when π₯ is equal to four.

Weβre now ready to find an expression for dπ¦ by dπ₯. We need to differentiate π of π₯ all raised to the power of one-half with respect to π₯. Weβll do this by using the general power rule. We set π equal to one-half and π of π₯ equal to π of π₯. This gives us one-half times π prime of π₯ multiplied by π of π₯ all raised to the power of one-half minus one.

And itβs worth reiterating, to use the general power rule, we needed that π of π₯ was differentiable. And in the question, the only value of π₯ we know π of π₯ is differentiable is π₯ is equal to four. So, the only time we can guarantee this formula will work is when π₯ is equal to four. Luckily, this is all we need to answer our question. Before we do this, we can simplify. To start, we can simplify the exponent one-half minus one to be equal to negative one-half. And then, weβll do one more piece of simplification. We know by using our laws of exponents, π raised to the power of negative one-half is the same as one divided by the square root of π. So, by using this, we can simplify our expression for dπ¦ by dπ₯ to be equal to π prime of π₯ divided by two times the square root of π of π₯.

And now, weβre ready to find dπ¦ by dπ₯ when π₯ is equal to four. We just need to substitute π₯ is equal to four into this expression for dπ¦ by dπ₯, and we know that this is valid when π₯ is equal to four. Substituting π₯ is equal to four into this expression, we get dπ¦ by dπ₯ at π₯ is equal to four is equal to π prime of four divided by two times the square root of π of four. And, of course, in the question, weβre told π prime evaluated at four is equal to two and π evaluated at four is equal to seven. So, weβll substitute these into this expression.

So, substituting π prime of four is equal to two and π of four is equal to seven, we get two divided by two times the square root of seven. And, of course, we can simplify. We can cancel the shared factor of two in our numerator and our denominator to give us one divided by the square root of seven. And we could leave our answer as one divided by the square root of seven. However, we can simplify this by rationalizing our denominator. Weβll multiply both our numerator and our denominator by the square root of seven. And this leaves us with our final answer of the square root of seven divided by seven.

Therefore, we were able to show if π¦ is equal to the square root of π of π₯, π prime evaluated at four is equal to two, and π of four is equal to seven, then dπ¦ by dπ₯ at π₯ is equal to four must be equal to the square root of seven divided by seven.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions