Derivatives of Exponential and Natural Log Functions We will introduce the concept of early transcendentals now, and then you will learn more about them as we go along. We'll just incorporate them into our existing knowledge of products, quotients and derivatives. Now we will examine the derivatives of e to the x and natural log of x. Let's begin by doing e to the x. There are many ways to prove this. One definition of e is 1 plus x plus x squared over 2 factorial plus x cubed over 3 factorial plus x to the fourth over 4 factorial. This will go on forever. To find the derivative of e to the x, we need to take the derivative of both sides of the equation. Over here we do not yet know how to take the derivative of e to the x, but we can take the derivative of this term since it is just a polynomial. The derivative of 1 is 0. The derivative of x is 1 plus the derivative of x squared, which is 2x divided by one of the 2's down here, would give us x plus the derivative of x cubed, which is 3x squared divided by 1 of the 3's here, would give a 2. So, that's 2 factorial. Then, by factoring x out of the equation, we would get rid of one of these 4s. So this would equal x cubed over 3 factorial. And this goes on forever. Notice that this is exactly what we started with: the definition of e to the x. The derivative of e to the x is its own value. The slope at any point on the graph of e to the x is also just e to the x. There are a variety of ways to arrive at this conclusion, but the slope of this curve is itself, for any value of x. I will take the derivative of ln x now. I want to define y as equal to ln x. Since I do not know the derivative yet, I will not take it, but instead exponentiate both sides of the equation, which will result in exponents of a base e.
So, if we let e to the y equal e to the x, then by the properties of logs, we can see that x is equal to e to the y. Now, in order to solve for e to the y, we differentiate both sides with respect to x. Once this derivative is done, we know the derivative of e to the x is e to the x, so derivative of e to the y is e to the y. But by implicit differentiation I multiply it by dydx - the derivative of function y with respect to x. The derivative of x is 1. We will solve for dydx by dividing both sides by e to the y. The derivative of our function y, or the derivative of natural log x, is equal to 1 over e to the y. This is because earlier we defined e to the y to be equal to x. Therefore, we see that the derivative of the natural logarithm with respect to x is 1 over x, which is the derivative of y with respect to x. This is the definition. The derivative of the natural logarithm of x is 1/x. As you can see, the derivatives of exponential and logarithmic functions are simply another instance where memorizing formulas is necessary, as we do in later chapters.