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 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.
