Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
By the product rule, the derivative of a product of functions f(x)g(x) is f(x)g'(x) + f'(x)g(x). So, in order to calculate the derivative of a product of functions, calculate the derivatives of the individual functions. Then, simply multiply the derivative of the second function to the first function, and add the derivative of the first function multiplied by the second function.
For example, if your function h(x)=x2ln(x), let f(x)=x2 and g(x)=ln(x). We know by the rules for deriving power functions and logarithmic functions that f'(x)=2x and g'(x)=1/x. h'(x)=f(x)g'(x) + f'(x)g(x). So, h'(x)=x2(1/x) + 2xln(x) = x + 2xln(x).
So we're talking about the product rule. You recall that the product rule says that the derivative of the product of two functions, let's say f, and g is the first function times the derivative of the second, plus the second function, times the derivative of the first.
It's a little easier to write this way using the prime notation. So you have f(x) times g(x) the quantity prime equals f(x) times g'(x) plus g(x) times f'(x). It still says the same thing. First times derivative of the second, plus second times derivative of the first.
Let's use this in an example. So I'm asked to differentiate this product. I want dy/dx, and that's going to be the derivative of this guy here. So it will be the first times the derivative of e to the x, plus the second e to the x times the derivative of x² minus 2x minus 3.
This is just going to be e to the x. So x² minus 2x minus 3 times e to the x. Then this is e to the x times some polynomial, 2x minus 2. Now you'll notice that both of these terms have an e to the x factor. So they probably simplify if I put them together. How many e to the x's do I have? I have x² of them minus 2x here plus 2x here. That just ends up being 0. Then minus 3 minus 2, so minus 5 times e to the x. So that's my final answer.
Then in part b I'm asked to differentiate this product; dy/dx equals again it's first times the derivative of the second. So x times the derivative of lnx plus lnx times the derivative of x. The derivative with respect to x of x.
Now recall that the derivative of lnx is 1 over x. So this is x times 1 over x. Then we have lnx times, and the derivative of x, well x is just a linear function with slope 1. The derivative of the linear function is at slope, so this is just 1. This gives me 1 plus lnx. So the derivative of x lnx is plus lnx. That's it.
Remember the product rule. The first times the derivative of the second plus the second times the derivative of the first.
