Special Derivatives

Proof: Derivative of sin(x)

This is a really interesting proof, and one that we can do easily, so let’s give it a shot. Since we can’t use the power rule here, let’s go back to limits, the original definition of a derivative, and plug in our function sin(x), giving us:

$\frac{d}{dx}\sin \left(x\right)=\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin \left(x\right)}{h}$

Using the angle addition trigonometric formula found here, we can expand the numerator to

$=\underset{h\to 0}{lim}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\cos h+\cos x\sin h-\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}{h}$ ​
notice that the two sin(x) parts can be taken out using the distributive property, giving us

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

Now let’s split the numerator into two separate limits. Can you see where this is going?

$=\underset{h\to 0}{lim}\frac{\sin x(\cos h-1)}{h}+\underset{h\to 0}{lim}\frac{\cos x\sin h}{h}$​
How about now? There should be something familiar in these two limits to you. Perhaps rearranging it might help.

$=\sin x\underset{h\to 0}{lim}\frac{\cos h-1}{h}+\cos x\underset{h\to 0}{lim}\frac{\sin h}{h}$

I’m feeding you the answer now! These are two of the most important limits we learned earlier! You should be able to see that the first term approaches 0 and the second term approaches 1. Refer to Gallery 2.3.1 if you are unsure.

Once we evaluate the two limits, we can see that the proof simplifies to

$=\sin x\cdot (0)+\cos x\cdot (1)$

$=\cos x$

Finally, we prove that $\frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathbf{sin}\mathit{x}\mathbf{=}\mathbf{cos}\mathit{x}$

Proof: The derivative of e^x

^{$\frac{d}{dx}{e}^x=\underset{h\to 0}{lim}\frac{e^{x+h}-e^x}{h}$​}
Using exponent laws, we can rearrange part of the numerator

$\underset{h\to 0}{lim}\frac{e^x\cdot e^h-e^x}{h}$​
Writing it this way allows us to factor out the bolded e^x.

$=\underset{h\to 0}{lim}\frac{{\mathbf{e}}^{\mathbf{x}}(e^h-1)}{h}$

We can take the e^x out of the limit since it is independent of h, giving

After factoring out e^x, the remaining limit is $\underset{h\to 0}{lim}\frac{e^h-1}{h}$. For the exponential with natural base e, this limit equals 1. We use this established exponential limit; we cannot replace e by a finite-h expression just because that expression tends to e.

$\frac{d}{dx}{e}^x=e^x\cdot \underset{h\to 0}{lim}\frac{e^h-1}{h}$

The key step is a limit, not cancellation of a substituted expression: lim_h→0 (e^h − 1)/h = 1.

Using that exponential limit, we obtain d(e^x)/dx = e^x · 1.

$\frac{d}{dx}{e}^x=e^x$

The derivative of e^x is e^x. Its slope and its height agree at every real x—a pretty useful property!

More generally, every function C e^x, with C constant, has derivative equal to itself. This includes the zero function when C = 0.

Usually, the first step in these problems is to convert everything into functions of sine and cosine. However, since this problem is already converted, let’s proceed to splitting the numerator and taking the derivative. We can see that another way to write our function is

$=\frac{d}{dx}(\frac{\cancel{\sin x}}{\cancel{\sin x}\cdot \cos x}+\frac{\cancel{\cos x}}{\sin x\cdot \cancel{\cos x}})$

$=\frac{d}{dx}(\sec x+\csc x)$

$=\sec x\tan x-\csc x\cot x$

Let’s simplify the terms into sin(x) and cos(x)​
$=\frac{d}{dx}(\cos x+\sin x\cdot \frac{\sin x}{\cos x})$

Now we have two separate terms that we cannot add together because they do not have a common denominator (yet). So let’s make it so that they do have a common denominator.​
$=\frac{d}{dx}(\frac{{\cos }^{2}x}{\cos x}+\frac{{\sin }^{2}x}{\cos x})$

You can probably see where this is going now. After adding the terms and simplifying using our trig identities, we get

$=\frac{d}{dx}\sec x$

$=\sec \left(x\right)\cdot \tan \left(x\right)$