### 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.

Weβre 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, weβll 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 weβre taking the square root of the product of two functions. This means weβre 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, weβll use the general power rule.

To start, letβs call π of π₯ three π₯ times π to the power of negative four π₯ plus four. Thatβs the entire function inside the square root symbol of π¦. So by using this definition for our function π of π₯, weβve 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 itβs the product of differentiable functions. So to find the derivative of π¦ with respect to π₯, weβll use the general power rule.

So letβs 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 πth 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 weβre 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, weβre going to need to use the product rule. So letβs 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₯. Thatβs the derivative of three π₯ times π to the power of negative four π₯ plus four with respect to π₯.

First, weβll evaluate this derivative term by term. Thatβs 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 weβll 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, weβre going to need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs 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.

Letβs now find an expression for π£ prime of π₯. Thatβs the derivative of π to the power of negative four π₯ with respect to π₯. To do this, weβre 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 ππ₯ with respect to π₯ is equal to π times π to the power of ππ₯.

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₯. Weβre now ready to use the product rule to find an expression for π prime of π₯. In fact, we couldβve 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 weβll 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, thereβs 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 weβll 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 weβll 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.