Chain Rule: The General Exponential Rule - Problem 3
We’re differentiating functions of the form e to the g(x). And in this next example we have a function h(x) equals 50 over 1 plus 4e to the -0.05x. Now here, our e to the g(x) is a part of a larger function. So we’re going to end up differentiating this somewhere along the way. But notice, the overall, this is a quotient of 2 functions.
Functions of this form, where you have a constant over 1 plus and then you have an exponential decay function, this is called a logistic function. It’s very important in modelling population growth. Let’s take a look at its derivative.
We’re going to need as the quotient rule here. We have h'(x) equals, and then remember the quotient rule. It says low d high, so 1 plus 4e to the -0.05x, d high would be the derivative of 50 which is 0, minus high d low 50 times the derivative of the bottom. The derivative of the bottom is going to be; well the derivative of the 1 plus part is going to be 0. And the derivative of the 4e to the -0.05x, we have 4e to the -0.05x times the derivative of this part. So we write that down. 4e to the -0.05x times the derivative of this part here, and that’s going to be -0.05. And then we write (1 plus 4e to the -0.05x)².
This is over the square of what’s below, if you recall the quotient rule. Now this term completely zeros out. Over here, we’ve got -50 times 4 times -0.05. -50 times 4 is -200. The negatives are going to cancel, we get positive 200. Now 0.05 is the same as 1/20. So 200 times 1/20 is going to be 10. We end up with 10e to the -0.05x in the numerator and in the denominator, we have (1 plus 4e to the -0.05x)² and that’s our final answer.