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Chain Rule: The General Exponential Rule - Problem 3

Teacher/Instructor 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.

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, quotient 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 quotient rule states that the derivative of h(x)=f(x)/g(x) is h'(x)=(g(x)f'(x) - f(x)g'(x))/(g(x))2.

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.

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