 ###### 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.

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# Chain Rule: The General Exponential Rule - Problem 2

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.

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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, product rule, and rules for exponential functions. So, when finding the derivative of some exponential function involving a composite function, use the chain rule to find the derivative of the composite part, and then derive the exponent as you normally would. Recall that the derivative of ex is itself, ex. Also, remember that the product rule states that the derivative of h(x)=f(x)g(x) is h'(x)=f(x)g'(x)+f'(x)g(x).

We’re talking about differentiating function of the form e to the g(x). Sometimes, that function’s not the headlining function. You have h(x) in this case equals x squared minus 3 times e to the negative x.

This is a form of e to the g(x) but it’s not the main event here. We have to use the product rule to deal with a function of this type, because, the first thing that we notice is,this is a product of x² minus 3 and e to the minus x. But we will need to use this rule when we differentiate e to the minus x. Let’s see how that works.

I’ve got the function written up here, so h' will be the product rule. First x² minus 3 times the derivative of the second and by the general exponential rule, it’s e to the -x times -1. The derivative of –x. Plus since under the product rule, e to the –x times the derivative of the first and that’s going to be just 2x.

Now let’s take a look at what we have here. I can factor out a -e to the –x. And I’m left with x² minus 3 and also if I factor –e to the –x out of this, I’m left with -2x. That’s my final answer. The derivative is –e to the –x, times the quantity x² minus 2x minus 3.