# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 8

if π(π₯) = 4π₯Β² + 6 + 2 ln 4π₯, find πβ²(2).

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### Video Transcript

if π of π₯ is equal to four π₯ squared plus six plus two ln four π₯, find π prime of two.

π prime is the derivative of our function with respect to π₯. So weβre going to need to differentiate four π₯ squared plus six plus two ln four π₯. Letβs do this by individually differentiating four π₯ squared, six, and two ln four π₯ with respect to π₯. Recall the derivative of ππ₯ to the power of π, where π is a real number and π is a constant, is πππ₯ to the power of π minus one. And this means we can differentiate four π₯ squared with respect to π₯. Itβs two times four π₯ to the power of one.

Well, two times four is eight and π₯ to the power of one is just π₯. So the derivative of four π₯ squared with respect to π₯ is eight π₯. When we differentiate a constant such as the six here, we get zero. But what about the derivative of two ln of four π₯? Well, we know that for constant values of π, the derivative of ln ππ₯ with respect to π₯ is one over π₯. And we use the constant multiple rule. This says that we are allowed to take a constant outside of a derivative and concentrate on differentiating the function of π₯ itself.

So the derivative of two ln of four π₯ with respect to π₯ is two times one over π₯ which is two over π₯. And so, we can see that π prime of π₯ here is eight π₯ plus zero plus two over π₯, which of course we can write as eight π₯ plus two over π₯.

Now, weβre not quite finished. Weβre looking to find π prime of two. So we substitute π₯ is equal to two into our expression for the derivative. And that gives us eight times two plus two over two. Eight times two is 16 and two divided by two is one. 16 plus one is 17.

So if π of π₯ is equal to four π₯ squared plus six plus two ln of four π₯, π prime of two is 17.