Question Video: Differentiating Compositions of Exponential and Root Functions Using the Chain Rule | Nagwa Question Video: Differentiating Compositions of Exponential and Root Functions Using the Chain Rule | Nagwa

# Question Video: Differentiating Compositions of Exponential and Root Functions Using the Chain Rule Mathematics • Third Year of Secondary School

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Find the derivative of the function π¦ = β(3π₯π^(β4π₯) + 4).

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### Video Transcript

Find the derivative of the function π¦ is equal to the square root of three π₯ times π to the power of negative four π₯ plus four.

Weβre given π¦ which is a composition of many functions. We need to find the derivative of π¦. In fact, since π¦ is a function in π₯, we need to find the derivative of π¦ with respect to π₯. To start this question, weβll need to look at what type of function π¦ is. We can see inside of the square root sign, we have the product of two functions three π₯ multiplied by π to the power of negative four π₯ and then we add four. So weβre taking the square root of the product of two functions. This means weβre going to need to use the product rule and the chain rule.

However, we could also use our laws of exponents to rewrite π¦ as three π₯ times π to the power of negative four π₯ plus four all raised to the power of one-half. And now, we can differentiate this by using both the product rule and the general power rule. We can do either of these two methods. In this video, weβll use the general power rule.

To start, letβs call π of π₯ three π₯ times π to the power of negative four π₯ plus four. Thatβs the entire function inside the square root symbol of π¦. So by using this definition for our function π of π₯, weβve rewritten π¦ as the square root of π of π₯, or by using our laws of exponents, π of π₯ all raised to the power of one-half. And of course, we know that our function π of π₯ is differentiable. This is because itβs the product of differentiable functions. So to find the derivative of π¦ with respect to π₯, weβll use the general power rule.

So letβs recall what the general power rule tells us. The general power rule states for a differentiable function π of π₯ and real constant 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.

In our case, we have the function π of π₯ which we know is differentiable. And our exponent value of π will be equal to one-half. So now, we can apply the general power rule to find the derivative of π¦ with respect to π₯. We get π¦ prime is equal to one-half times π prime of π₯ multiplied by three π₯ times π to the power of negative four π₯ plus four all raised to the power of one-half minus one. And we can simplify this slightly. In our exponent, one-half minus one is equal to negative one-half.

But weβre not done. To find an expression for π¦ prime, we now need to find an expression for π prime of π₯. And if we look at our definition of the function π of π₯, to do this, weβre going to need to use the product rule. So letβs now recall the product rule.

The product rule tells us if a function π of π₯ is the product of two differentiable functions π’ of π₯ times π£ of π₯, then π prime of π₯ is equal to π’ prime of π₯ times π£ of π₯ plus π£ prime of π₯ times π’ of π₯. And we can use this to find an expression for π prime of π₯. Thatβs the derivative of three π₯ times π to the power of negative four π₯ plus four with respect to π₯.

First, weβll evaluate this derivative term by term. Thatβs the derivative of three π₯ times π to the power of negative four π₯ with respect to π₯ plus the derivative of four with respect to π₯. Of course, four is a constant, so its derivative with respect to π₯ is equal to zero. So in actual fact, to find an expression for π prime of π₯, we only need to differentiate three π₯ times π to the power of negative four π₯. So weβll set our function π’ of π₯ to be three π₯ and π£ of π₯ to be π to the power of negative four π₯.

To apply the product rule, weβre going to need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs the derivative of three π₯ with respect to π₯. Three π₯ is a linear function, so its derivative with respect to π₯ will be the coefficient of π₯, which is three.

Letβs now find an expression for π£ prime of π₯. Thatβs the derivative of π to the power of negative four π₯ with respect to π₯. To do this, weβre going to need to recall one of our standard rules for differentiating exponential functions. For any real constant π, the derivative of π to the power of ππ₯ with respect to π₯ is equal to π times π to the power of ππ₯.

In our case, our value of π is equal to negative four. So we get π£ prime of π₯ is equal to negative four times π to the power of negative four π₯. Weβre now ready to use the product rule to find an expression for π prime of π₯. In fact, we couldβve just slightly changed our product rule to evaluate this from the first line of our expression. So the product rule tells us π prime of π₯ will be equal to π’ prime of π₯ times π£ of π₯ plus π£ prime of π₯ times π’ of π₯.

Substituting in our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯, we get that π prime of π₯ is equal to three times π to the power of negative four π₯ plus negative four π to the power of negative four π₯ multiplied by three π₯. And weβll simplify this expression by taking out the shared factor of three times π to the power of negative four π₯. This gives us three π to the power of negative four π₯ multiplied by one minus four π₯.

But remember, the question is asking us to find an expression for the derivative of π¦ with respect to π₯. So we need to substitute this expression for π prime of π₯ into our expression for π¦ prime. Substituting in our expression for π prime of π₯, we get π¦ prime is equal to one-half multiplied by three π to the power of negative four π₯ times one minus four π₯ multiplied by three π₯ times π to the power of negative four π₯ plus four all raised to the power of negative one-half.

And we could leave our answer like this. However, thereβs a little bit of simplification we can do. First, by using our laws of exponents, π raised to the power of negative one-half is equal to one divided by the square root of π. So weβll use this to move three π₯π to the power of negative four π₯ plus four all raised to the power of negative one-half into our denominator. This gives us three times π to the power of negative four π₯ multiplied by one minus four π₯ all divided by two times the square root of three π₯π to the power of negative four π₯ plus four.

And the last piece of simplification weβll do is rearrange our terms one and negative four π₯. And this gives us our final answer. Therefore, given the function π¦ is equal to the square root of three π₯π to the power of negative four π₯ plus four, we were able to use the product rule and the general power rule to find an expression for the derivative of π¦ with respect to π₯. We got this was equal to three times π to the power of negative four π₯ multiplied by negative four π₯ plus one all divided by two times the square root of three π₯π to the power of negative four π₯ plus four.

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