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Chain Rule: The General Power Rule - Problem 1
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You can use the chain rule to find the derivative of a polynomial raised to some power. 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." In the case of polynomials raised to a power, let the inside function be the polynomial, and the outside be the power it is raised to. Then, by following the chain rule, you can find the derivative.
For example, let f(x)=(x^{5}+4x^{3}-5)^{6}. This function would be very difficult to expand by hand in order to find the derivative, so the chain rule would make things simpler. The inside function would be x^{5}+4x^{3}-5, and the outside function would be x^{6} (raising some unknown thing to the 6th power). The derivative of some y^{6} is 6y^{5} by the power rule, so the first part of the derivative is 6(x^{5}+4x^{3}-5)^{5}. Then, the derivative of the inside function is 5x^{4}+12x^{2}. Therefore, the derivative of this function would be 6(x^{5}+4x^{3}-5)^{5}(5x^{4}+12x^{2}).

Let’s do a harder example of the chain rule. Now we have a special case of the chain rule. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). So I want to apply this to h (x) equals x minus 1 over x plus 1 all raised to the 3rd power.

First of all, let’s note that the inside function is x minus 1 over x plus 1. So if I were color-coding this, I would make this raising to the 3rd power, that would be the outside function. Put that in red. The inside function would be this rational function. Now just thinking ahead, what do I do to the inside function? I do have to differentiate it at the end here. So at some point I am going to have to use the quotient rule to differentiate this function.

Let me begin. H'(x) is going to be, now differentiating the outside function first, the something to the 3rd power part in the function. And the derivative of that is 3 times something to the 3 minus 1. 3 times something squared. I don’t touch the inside part, just leave it as it is; x minus 1 over x plus 1. Then times, and I want the derivative of the inside function. Here is where I use the quotient rule. The quotient rule says low d high, x plus 1 times the derivative of the top minus high d low, the (x minus 1) times the derivative of the bottom over the square of what’s below. x plus 1 is in the bottom so I square that.

This will simplify a lot. Let me take this up over here. So I have h'(x) equals: what do I have so far? I’ll just this black and make the color-coding more. 3((x minus 1) over x plus 1)². Let me go back really quickly.

This numerator is going is going to simplify. I have an x plus 1 and an x minus 1. I’m subtracting x minus 1. So the xs are going to cancel. X minus x, that will cancel and I have 1 minus -1, so 1 plus 1. That gives me a 2 in this numerator. And then the x plus 1 squared in the denominator. So let me write that down, 2.

Now looking at this term, notice, I can distribute the square over the numerator and denominator. I would have (x minus 1)² on top and (x plus 1)² on the bottom. I’m multiplying that by 3 and by 2 over (x plus 1)². Right now I’m just simplifying. On the top I’m going to have 3 times 2, 6 times (x minus 1)² an in the bottom I have (x plus 1) to the 4th power. That’s my derivative, h’(x).

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