GodsMathProject
- 1. Julie Metzler and Jesus Benitez Final Project We will be discussing the function ex and its unusual derivative: itself. This is strange because that means that at each x value, the slope is equal to the y value. We will prove why this is the case using limits and a graph, implicit differentiation, and even the chain rule and the derivative of the natural log. First, we proved that the derivative of ex is itself using the limit definition of a derivative. Which is: So we plugged ex into the equation to get: Because of the laws of exponents, we split the addition of the x and h exponents into the multiplication of base e to the powers of e and x. Next we factored an ex out of the numerator. Then we put the ex in front of the limit because, like other derivatives and integrals, the expression is a constant, and therefore can be added into the problem after, in this case, the limit has been taken. Here we can see that as h approaches 0, the limit will get closer to 0/0. This is an indeterminant form because both the numerator and denominator approach 0. Though this does not yield an exact answer, we plugged into the equation the point (0,1) to observe the function’s behavior at this point.
- 2. (Graph compliments of www.wyzant.com) Because this limit definition, we found that e has the limit of 1 when going to zero, which is shown on the graph, and here in this equation. Therefore, we can substitute 1 for the limit in our limit equation, making this: into this: This shows that though the limit going to 0 is indeterminate because of its 0/0 form, the derivative of ex is actually itself. Another way that we decided to prove that the limit of ex is really itself is using implicit differentiation using the natural log of x.
- 3. We let ex=y so the ln(ex)=ln(y), and thus making x=ln(y). Next we took the derivative of y with respect to x to get: Next, we multiplied both sides by y to get y alone. Then we substituted ex for y because that was our original value for y. This implicit differentiation clearly shows that ex is equal to its derivative. Lastly, we addressed the chain rule and the derivative of the natural log, which as we all know is the opposite of ex, to completely ensure that our calculations above were not flawed. We let f(x)=ln(u), and u=ex to get the derivative of ln(ex). This gave us the equation: Since we know that the derivative of the natural log is 1 divided by whatever you are taking the natural log of, we know that ln(u)=1/u. We also know that the ln(ex) is 1. Because of this, we substituted 1/u for d/du ln(u) and 1 for d/dx ln(u). This left us with the equation: Next we multiplied both sides of the equation by u to get it alone and out of the denominator to get: Lastly we substituted ex back in for u. This left us with the familiar equation: We saw this same equation when we took the derivative of ex using implicit differentiation.
- 4. We proved that the rule that the derivative of ex is itself is valid. To be sure of our calculations, we used three different methods and a graph to prove the rule and to ensure that our calculations were not skewed. The function ex is truly an unusual one, but extremely effective in the calculations of derivatives and integrals. Special thanks to our professor Mr. Keith Nabb for inspiring us to want to do all this work just for the fun of it, and to Mathaway.com for making it easier to type in our equations in the format we wanted and allwoing us to copy them here. They have singlehandedly made our project so much neater and more professional looking.