Derivatives Of Exponentials, Sine and Cosine

Derivatives of Exponential Functions

Recall that an exponential function is a function of the form $f(x)=a^x$ 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 $y=2^x$ represents repeated doubling. What’s its derivative?

$$\lim_{h\to 0}\frac{2^{x+h}-2^x}{h}$$

But using the laws of exponents, $2^{x+h}=2^x2^h$ so we can factor $2^x$ out of the numerator, and then pull it out front of the limit (its a constant since the limit is only about the variable $h$)! $$\frac{d}{dx}2^x=2^x\lim_{h\to 0}\frac{2^h-1}{h}$$ 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 $2^x$ is just some *constant multiple of $2^x$! What is the constant?

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

We can actually plot the function $(2^h-1)/h$ and zoom in on $h=0$ to attempt to evaluate the limit:

The same thing works for $3^x$, 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 $e^x$

When is this proportionality constant equal to $1$? Can we solve for the base $a$ where $$\lim_{h\to 0}\frac{a^h-a}{h}=1?$$

When $a\simeq 2.71$ it appears that its slope at the origin is 1, and so $$\frac{d}{dx}a^x=a^x$$ This number is so important in calculus that it gets its own name, $e$, after its discoverer Euler:

$$e=2.718281828459\ldots$$

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:

$$(\sin x)^\prime = \lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}$$ Using the identity $\sin(x+h)=\sin(x)\cos(h)+\cos(x)\sin(h)$, we can simplify this as

$$\lim_{h\to 0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}$$

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

$$\lim_{h\to 0}\frac{\sin(x)\cos(h)-\sin(x)}{h}+\lim_{h\to 0}\frac{\cos(x)\sin(h)}{h}$$

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

$$\sin(x)\lim_{h\to 0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h\to 0}\frac{\sin(h)}{h}$$

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!

$$\frac{d}{dx}\sin(x)=0\sin(x)+1\cos(x)=\cos(x)$$

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, $$\frac{d}{dx}\sin(x-\pi/2)=\cos(x-\pi/2)=-\sin(x)$$