The Derivative of (e^x) As a follow-up to Carter Bowles, “Derivatives, An Introduction to Calculus, we will learn how to differentiate the natural exponential function (e). What are the applications of the natural exponential function? Well it is vital in the prediction of exponential growth and decay, along with the natural logarithmic function represented by (ln). Another application of (e) is closer to a physicist’s heart; it is Newton’s Law of Cooling. Yet another application of log’s and exponential functions is in Up and down motion in a resisting medium and motion in one direction in a resisting media. The natural logarithm (ln) and log will be discussed in a later article. Now, let us see a couple examples of these very simple exponential functions. An example of a natural exponential function may be this: f(x) = 4e2x+x² Now how can we go about finding (dy/dx) for the previous function 4e2x+x² ? We follow a simple rule for differentiating natural exponential functions. Let f(x) be differentiable for x, for f(x) = eu , let f `(x) = u`eu. So we can now go back to our original equation and find the derivative of f(x). So, To find u` we can use our knowledge of derivatives and simple power rule: u`= 2 + 2x Therefore, f `(x) = (2 + 2x) (4)e2x+x² = (8 + 8x) e2x+x² Here we have our first Natural exponential derivative. The same rules for regular derivative apply here as well. Take this function for example: y = xex , this is a very simple natural product rule. If we remember our product rule theorem it states that for a two functions f(x) and g(x), both differentiable for x, that if f(x)∙g(x), the derivative is f(x) ∙ g`(x) + f `(x) ∙ g(x). Now we can apply this product rule to the function y = xex. y = x∙ex , therefore we can apply our product rule for differentiation. The Derivative of (e^x) #2 dy/dx = x∙ex + 1∙ ex = xex + ex There we had a very easy differentiation of the natural exponential function. Let us now consider a more difficult example encompassing all our previous knowledge. Here is the problem: Find the derivative of the function f(x) = (x2 – 4) e√x f(x) = (x2 – 4) ex^(1/2) u = ex^(1/2) u` = ½ (x)-1/2 (1) (chain rule) u` = 1/2√x Now once we find u` we can continue on with product rule: f `(x) = (x2 – 4)∙ e√x/2√x + (2x) e√x So that is our final answer: f `(x) = (x2 – 4)∙ e√x/2√x + (2x) e√x This problem encompassed not only product rule, but chain rule, and our knowledge of the natural exponential function e. Some people may say combine the terms into a single numerator and denominator, but to get the idea of the derivation of e this doesn’t matter. This method works for any exponential function not just those with base (e). I hope this straightforward approach helps you understand just a little more of the Calculus.