Derivatives
A derivative just describes how much one quantity changes as you vary another quantity that it depends on. Graphically, if you have a curve, the derivative at a point gives the slope of the line touching the curve (tangent) at that point. Here we'll just describe how to compute most of the cases you'll ever need.
Formula
$$\frac{d}{dx}x^n=nx^{n-1}$$
Proof
Use definition of derivative as the limit of a difference:
$$\frac{d}{dx}x^n=\lim_{\Delta x\to 0}\frac{(x+\Delta x)^n-x^n}{\Delta x}.$$
Since we are taking the limit as $$\Delta x\to 0,$$ we can treat it as very small compared to $$x,$$ and rewrite the above as
$$\lim_{\Delta x\to 0}\frac{x^n(1+\Delta x/x)^n-x^n}{\Delta x}\approx \lim_{\Delta x\to 0}\frac{n x^n\Delta x/x}{\Delta x}$$
$$=nx^{n-1}.$$
The approximate equality comes from using the "most useful approximation ever" we showed in the Basic Math Review:
$$\big[1+\epsilon\big]^n\approx 1+n\epsilon.$$
Formula
$$\frac{d}{dx}e^x=e^x$$
Proof
Write
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ and differentiate each term using the previous formula to find
$$\frac{d}{dx}e^x=\sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}+0$$
where the 0 comes from the $$n=0$$ term in the original series, which has derivative zero.
Then re-index the sum to again start at 0, by transforming $$n-1\;\rm{to}\;n.$$ The result then follows.
Note: the formula for e as a sum above is super-useful!
It immediately gives you the approximation for e at small x by just dropping terms that are quadratic or above in x:
$$e^x\approx 1+x$$
as well as an approximation for $$\ln\big[1+x\big]$$
by taking the natural log of both sides of $$e^x\approx 1+x$$
to get
$$x\approx \ln\big[1+x\big]$$ for small x.
Incidentally, the series for e can also be used, by comparison with the series expansions for sine and cosine, to prove Euler's formula! (substitute in $$x=i\theta$$ and just algebra it out!)
Chain Rule
A very useful way to do derivatives of compositions of functions. What is a composition of functions? Well, $$\big[1+x\big]^2$$ is one example! Inside the brackets, you take x to 1+x. That's one function:
$$f(x)=1+x.$$ Then outside the brackets, you take that function, f, to its square: $$g(f)=f^2.$$
So we can write the initial expression as g of f. What if we want to do the derivative of something of that form? It is
$$\frac{d}{dx}\big[g\circ f\big]=\frac{dg}{df}\frac{df}{dx}$$
An easy way to remember this is to imagine the df's canceling out, much like how one would write out unit conversions, e.g. meters to feet as $$1\; \rm{m} \times \frac{3\;\rm{ft}}{\rm{m}}=3\;\rm{ft}.$$ We won't prove this formula here---maybe you can think of a proof!
Formula
$$\frac{d\ln x}{dx}=\frac{1}{x}$$
Proof
Write x as
$$x=e^{\ln x},$$ which is true by the definition of ln. Then differentiate both sides, using Chain Rule for the righthand side:
$$1=e^{\ln x}\frac{d\ln x}{dx}.$$
Then simplify using that $$e^{\ln x}=x,$$ and divide both sides by x:
$$\frac{1}{x}=\frac{d\ln x}{dx}.$$
Formula
$$\frac{d\sin x}{dx}=\cos x,\;\frac{d\cos x}{dx}=-\sin x.$$
Proof
Use Euler's formula to write sin and cos as complex exponentials.
Then
$$\frac{d\sin x}{dx}=\frac{d}{dx}\bigg(\frac{1}{2i}\big[e^{ix}-e^{-ix}\big]\bigg)=\frac{1}{2i}\big[ie^{ix}+ie^{-ix}\big].$$
The second equality above is using Chain Rule to differentiate $$e^{ix}.$$
Simplifying the righthand side, we have
$$\frac{1}{2}\big[e^{ix}+e^{-ix}\big]=\cos x.$$
It's nearly the same to compute the derivative of cos: try it for yourself!
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