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 1

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 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.

A lot of really simple exponential functions fall under the category of general exponential function, which I call functions of the form e to the g(x). I'm going to deal with a bunch of examples that are e to the, and then constant times x, because they’re all differentiated the same way. We use the general exponential rule, but it will be a really simple version of it. So, when you’re differentiating something like e to the -2x, the derivative is e to the -2x, times the derivative of -2x.

The derivative of -2x is -2. So you’re just going to get e to the -2x times -2, or -2e to the -2x. And this is something you’ll see a lot. So let’s take a look at three more examples that are kind of like this. The derivative of e to the 0.05x. Same thing. We have like a little m times x here and the derivative is going to be e to the 0.05x times the derivative of 0.05x which is 0.05.

And then here just in general, if you just have a value mx, it’s going to be e to the mx times m. So you can think of this as a very special formula for functions of the type e to the constant times x. Where does this function fit in? it turns out that functions of they type a to the x, can always be written as e to the m times x. 2, an identity that involves exponential functions, 2 can be written as e to the ln2.

Any number 3 is e to the ln3. So you can always use this identity to rewrite this function, with base e. So you’d get the derivative, this is the same as the derivative of e to the natural log 2 times x. These are the same function.

So when I differentiate this function keeping in mind that ln2 is just a constant, I get e to the ln2 times x times ln2. And because this is the same as 2 to the x, I can write 2 to the x times ln2. And you recall that this is the formula that we had before we differentiated 2 to the x. This is how you can derive the formula. Just change the 2 to e to ln2, and that’s how you differentiate any exponential function.

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