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 derivative of e^{x} is itself, e^{x}. So, by using the sum rule, you can calculate the derivative of a function that involves an exponential term. For example, let f(x)=7x^{3}-8x^{2}+2+4e^{x}. By using the power rule, the derivative of 7x^{3} is 3*7x^{2}=21x^{2}, the derivative of -8x^{2} is 2*(-8)x=-16x, and the derivative of 2 is 0. Then, using what we know about the derivative of e^{x}, we know that the derivative of 4e^{x} is simply 4e^{x}. So, the derivative of the entire function is f'(x)=21x^{2}-16x+4e^{x}.

Let’s try a couple of harder problems. We are asked to differentiate each function. First we have f(x) equals 2e to the x plus x². So f' would be the derivative and that’s going to be the derivative with respect to x of 2e to the x plus x².

Now I can separate these using the sum rule; the derivative of two e to the x plus the derivative of x². And then I can use the constant multiple rule to pull these two out. So I get 2 times the derivative of e to the x, plus the derivative of x² remember that it's 2x.

One of the big results of this section was the derivative of e to the x is just e to the x. So this is 2 times e to the x plus 2x. Let’s take a look at another one. G(x) equals 4 minus e to the 2 plus x. Now this one is tricky because it’s not clear exactly what to do with the e to the 2 plus x. It’s not e to the x. I need it to be e to the x. One thing I can do is rewrite this function as 4 minus. And then using properties of exponents, this would be e to the 2 times e to the x. E to the 2, e². This is just a constant, so I differentiate using constant multiple rule to pull out. So let’s do that.

The derivative g'(x), is going to be the derivative with respect to x of 4 minus e² times e to the x. I will separate this using sum rule. Get the derivative with respect to x of 4 plus the derivative with respect to x of –e², e to the x.

The derivative with respect of 4, you can think of as a linear function 0x plus 4. But its slope is going to be 0, plus and then I can pull this constant. Let me actually just write minus e². I'm pulling this constant outside times the derivative of e to the x. And so I get minus e², times e to the x. Again this is just e to the x.

And if you like you can put this back in the original form that it came in. We can multiply these together and get e to the 2 plus x. And so that’s our final answer. Minus e to the 2 plus x.