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# Chain Rule: The General Power Rule - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Remember that the chain rule is used to find the derivatives of composite functions. First, determine which function is on the "inside" and which function is on the "outside." The chain rule will be the derivative of the "outside" function multiplied by the derivative of the "inside" function. The chain rule can be used along with any other differentiating rule learned thus far, such as the power rule and the quotient rule. So, when finding the derivative of some quotient involving a composite function, use the chain rule to find the derivative of the composite part, and then use the quotient rule as you normally would: the derivative of h(x)=f(x)/g(x) is h'(x)=(g(x)f'(x) - f(x)g'(x))/(g(x))^{2}.

Let’s do a harder example. I want to differentiate h(x) equals 64x to the 6th plus 2 over the quantity x to the 4th plus 1 to the 6th power. Let me use the general power rule but I’m only going to use it when I’m differentiating the denominator. But overall this function’s a quotient. So I’m going to need to need the quotient.

Let’s get started with the derivative. So h'(x) equals low d high, so I have x to the 4th plus 1 to the 6th, times the derivative of the numerator. That’s going to be 6 times 64, 384x to the 5th. Minus high d low, that’s going to be 64, x to the 6th plus 2 times the derivative of the denominator. For the derivative of the denominator I do need to use the general power rule. So the 6 is going to come in front, I have 6 (x to the 4th plus 1) and the exponent drops by 1 so 5. I have to multiply by the derivative of x to the 4th plus 1. That’s 4x³.

In the denominator, I get the square of what’s below, so the x to the 4th plus 1 to the 6th power becomes to the 12th power. And this is the raw form of the derivative; I just need to do a lot of simplification.

First let’s observe that this x to the 4th plus 1 appears in both of the terms of the numerator and the denominator. So some cancellation is going to take place. Let me do that with a different color. I have 5 of them here, 6 of them here and 12 down here. So I can cancel 5 between numerator and denominator. This one’s completely gone, 5 of these are gone, leaving one. And then 5 these are gone leaving 7.

I distribute this 384x to the 5th over the one remaining factor. I get 384x to the 9th plus 384x to the 5th. That’s take care of this part. Over here this factor’s now gone. I have 24x³ on the outside and I have to distribute this over these two terms. That’s all I have left 24x³.

24 times 64 is 1536. This is going to be minus, -1536x to the 9th. And then I have 2 times 24x³ and that’s going to be negative. So -48x³ all over x to the 4th plus 1 to the 7th power. All I have to do is combine like terms and I’ll be done. So 384x to the 9th minus 1536x to the 9th is -1152x to the 9th. I have an x to the 5th term, 384x to the 5th and -48x³. All that over (x to the 4th + 1) to the 7th power.

There’s a lot of simplification but usually your teacher will want you to simplify answers like this. Take a look at this function, it actually looks like Batman if you graph it out. Here is its complicated derivative; -1152x to the 9th plus 384x to the 5th minus 48x³ all over x to the 4th plus 1 to the 7th power.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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