Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part A β€’ Question 24

If 𝑓(π‘₯) = 2^(πœ‹π‘₯), what is the value of 𝑓′(π‘₯)?

02:21

Video Transcript

If 𝑓 of π‘₯ is equal to two to the power of πœ‹π‘₯, what is the value of 𝑓 prime of π‘₯?

Whenever we’re 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.

Let’s rewrite our function as 𝑦 is equal to two to the power of πœ‹π‘₯. And now, we can take the natural logarithm of both sides of this equation. When we do, we see that ln of 𝑦 is equal to ln of two to the power of πœ‹π‘₯. And here, we recall a law of logarithms that’s going to help us rewrite this further. This says that log of π‘Ž to the power of 𝑏 can be written as 𝑏 times log of π‘Ž. And since the natural logarithm is just a logarithm with a base of 𝑒, we can apply this rule to rewrite the right-hand side as πœ‹π‘₯ multiplied by ln two or πœ‹ ln two π‘₯.

And now we’ve 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 𝑦 with respect to π‘₯ by multiplying the derivative of the function with respect to 𝑦 by d𝑦 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 𝑦 with respect to π‘₯ is the derivative of ln 𝑦 with respect to 𝑦 which is one over 𝑦 times d𝑦 by dπ‘₯. And on the right-hand side, we remember that πœ‹ ln two is just a constant. And so, the derivative of πœ‹ ln two π‘₯ is πœ‹ ln two.

Now remember we’re being asked to find the value of 𝑓 prime of π‘₯ of the derivative of the function with respect to π‘₯. So we’re going to make d𝑦 by dπ‘₯ the subject. We multiplied both sides of our equation by 𝑦. And we see that d𝑦 by dπ‘₯ is equal to πœ‹ ln two times 𝑦.

Remember though we said that 𝑦 was equal to two πœ‹π‘₯. So we substitute this into our equation for the derivative. And we see that d𝑦 by dπ‘₯ and therefore our value for 𝑓 prime π‘₯ is πœ‹ ln two times two to the power of πœ‹π‘₯.

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