### Video Transcript

If π of π₯ is equal to two to the power of ππ₯, what is the value of π prime of π₯?

Whenever weβre finding the derivative of some constant with a variable exponent such as here, we need to do two things. We begin by taking the natural logarithm of both sides of our equation. And the reasons for this will become clear in a moment. We then use implicit differentiation to differentiate both sides of our equation.

Letβs rewrite our function as π¦ is equal to two to the power of ππ₯. And now, we can take the natural logarithm of both sides of this equation. When we do, we see that ln of π¦ is equal to ln of two to the power of ππ₯. And here, we recall a law of logarithms thatβs going to help us rewrite this further. This says that log of π to the power of π can be written as π times log of π. And since the natural logarithm is just a logarithm with a base of π, we can apply this rule to rewrite the right-hand side as ππ₯ multiplied by ln two or π ln two π₯.

And now weβve written it like this, we can use implicit differentiation to differentiate both sides of our equation with respect to π₯. Now, remember implicit differentiation is just a special version of the chain rule. And it says that we can find the derivative of this function in π¦ with respect to π₯ by multiplying the derivative of the function with respect to π¦ by dπ¦ by dπ₯.

Now remember the derivative of ln π₯ with respect to π₯ is one over π₯. And we can therefore say that the derivative of ln π¦ with respect to π₯ is the derivative of ln π¦ with respect to π¦ which is one over π¦ times dπ¦ by dπ₯. And on the right-hand side, we remember that π ln two is just a constant. And so, the derivative of π ln two π₯ is π ln two.

Now remember weβre being asked to find the value of π prime of π₯ of the derivative of the function with respect to π₯. So weβre going to make dπ¦ by dπ₯ the subject. We multiplied both sides of our equation by π¦. And we see that dπ¦ by dπ₯ is equal to π ln two times π¦.

Remember though we said that π¦ was equal to two ππ₯. So we substitute this into our equation for the derivative. And we see that dπ¦ by dπ₯ and therefore our value for π prime π₯ is π ln two times two to the power of ππ₯.