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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Derivatives of Exponential Functions - Problem 2

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 derivative of ex is itself, ex. So, by using the sum rule, you can calculate the derivative of a function that involves an exponential term. For example, let f(x)=7x3-8x2+2+4ex. By using the power rule, the derivative of 7x3 is 3*7x2=21x2, the derivative of -8x2 is 2*(-8)x=-16x, and the derivative of 2 is 0. Then, using what we know about the derivative of ex, we know that the derivative of 4ex is simply 4ex. So, the derivative of the entire function is f'(x)=21x2-16x+4ex.

Let’s talk about the derivative of a special exponential function e to the x. Now you recall that the derivative of an exponential function a to the x, is lna times a to the x. For the base e, this is going to be ln of e times e to the x. Of course Ln of e is 1 because e is the base of the natural log.

So this is the function whose derivative is itself. Derivative of e to the x is e to the x. That’s something very important about e to the x. Let’s take this and use in the problem. The problem says, to find an equation of the line tangent to the graph of f(x) equals e to the x at x equals 0 and sketch the tangent line.

If you want to find a tangent line you need two things. You need first, the point of tangency and second the slope. So let’s find the point tangency first. We know that we wanted the x equals 0, we are going to need f of 0 then. And that’s e to the 0. That’s just going to be 1.

So a point of tangency is x equals 0, y equals 1. Second, we need the slope and so we need the derivative of e to the x. Of course that’s just e to the x. And so the slope at x equals 0 is going to be e to the 0, which is 1. So that’s my slope.

And so to get the equation of the tangent, I just use point slope. Y minus the y coordinate, equals the slope times x minus the x coordinate. And that’s the equation on point slope. This can be simplified a lot if you put it in slope intercept. So this is going to be x and I’ve got plus 1. Y equals x plus 1.

So that’s the equation of the line tangent. Let me just show you what the graph of the line looks like. Here’s the graph of y equals e to the x. The point of tangency is 0,1, this point right here. So let me draw the line.

There is my tangent line right there. There we go. Y equals x plus 1, that’s the tangent to y equals e to the x at 0,1.

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