Derivatives Of Exponentials, Sine and Cosine

Derivatives of Exponential Functions

Recall that an exponential function is a function of the form f(x)=ax for some positive real number a, called the base.
For concreteness to start out, we will look at the base a=2, so the function y=2x represents repeated doubling. What’s its derivative?

limh02x+h2xh

But using the laws of exponents, 2x+h=2x2h so we can factor 2x out of the numerator, and then pull it out front of the limit (its a constant since the limit is only about the variable h)! ddx2x=2xlimh02h1h This tells us something incredible: *nothing* left in the limit depends on x, and so it is just a *constant*. That is, the derivative of 2x is just some *constant multiple of 2x! What is the constant?

It’s just the slope of 2x at x=0! We can estimate this by looking at the secant line

We can actually plot the function (2h1)/h and zoom in on h=0 to attempt to evaluate the limit:

The same thing works for 3x, or (3/4)x, etc: their derivatives are all just multiples of themselves!

What does this mean about exponentials? The rate at which they grow is proportional to the value they already have! That means, the “per-capita” rate of increase is constant! This is why they show up in physical modeling all the time.

The importance of ex

When is this proportionality constant equal to 1? Can we solve for the base a where limh0ahah=1?

When a2.71 it appears that its slope at the origin is 1, and so ddxax=ax This number is so important in calculus that it gets its own name, e, after its discoverer Euler:

e=2.718281828459

Derivative of Sine and Cosine

Use the angle sum rule for sin: sin(a+b)=sin(a)cos(b)+cos(a)sin(b) to evaluate the limit:

(sinx)=limh0sin(x+h)sin(x)h Using the identity sin(x+h)=sin(x)cos(h)+cos(x)sin(h), we can simplify this as

limh0sin(x)cos(h)+cos(x)sin(h)sin(x)h

We can break this into the sum of two simpler pieces:

limh0sin(x)cos(h)sin(x)h+limh0cos(x)sin(h)h

In the first of these we can factor out a sin(x) and in the second a cos(x):

sin(x)limh0cos(h)1h+cos(x)limh0sin(h)h

Just like for exponentials, we see that the two limits remaining do not depend on x. Their values are just numbers - so this already tells us a remarkable fact: the derivative of sine is made of just sines and cosines!

The value of the second of these limits is directly giving the slope of sin(x) at x=0:

This appears to be 1 from the graph, which we can confirm by drawing sin(h)/h *as a funciton of h:

The other limit measures the slope of cos(x) at x=0 (because cos(0)=1, this is just the slope of the secant line!)

Again we can confirm this by directly plotting the secant’s slope:

Becasue the first limit is 1 and the second is 0, this tells us that the derivative of sin(x) has a remarkably simple form!

ddxsin(x)=0sin(x)+1cos(x)=cos(x)

From this we can directly get the derivative of the cosine, recalling that cos(x) is nothing more than a shifted copy of sin(x):

Thus, ddxsin(xπ/2)=cos(xπ/2)=sin(x)