Basic
The Derivation of a function can be expressed as follows.
\[f'(x) = \lim_{\Delta x\rightarrow x_0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\]The formular for an power function is as follows.
\[f(x)=x^n\]Inserting the power function creates the following expression.
\[f'(x) = \lim_{\Delta x\rightarrow 0} \frac{(x+\Delta x)^n-x^n}{\Delta x}\]The binominal formular can be expanded.
\[f'(x) = \lim_{\Delta x\rightarrow 0} \frac{\frac{n!}{0!(n-0)!}x^n+\frac{n!}{1!(n-1)!}x^{n-1}\Delta x+\frac{n!}{2!(n-2)!}x^{n-2}\Delta x^2+...+\frac{n!}{n!(n-n)!}x^{n-n}\Delta x^n - x^n}{\Delta x}\]The first term result in \(\frac{n!}{0!(n-0)!}x^n = x^n\) and so it gets removed and so it becomes possible to perform the division with \(\Delta x\). Then inserting the limit \(\Delta x \rightarrow 0\) results in the following equation.
\[f'(x) = \frac{n!}{1!(n-1)!}x^{n-1} = nx^{n-1}\]Exponential
The formular for an exponential function is as follows.
\[f(x) = a^x\]Rewriting it enables a simple derivation
\[f(x) = a^x = e^{\ln(a)x}\]Derivation is as follows
\[f'(x) = \frac{d}{dx}a^x=\frac{d}{dx}e^{\ln(a)x}=\ln(a)e^{\ln(a)x}=\ln(a)a^x=\ln(a)a^x\]Logarithmic
The formular for an logarithmic function is as follows.
\[f(x) = \log_a x\] \[\frac{d}{dx}a^{f(x)} = \frac{d}{dx}x\] \[a^{f(x)}\ln a f'(x) = 1\]Inserting the function leads to as follows
\[a^{\log_a x}\ln a f'(x) = 1 = x\ln a f'(x)\] \[f'(x) = \frac{1}{x\ln a}\]Inserting \(a = e\) results in the following for \(f(x) = \ln x\)
\[f'(x) = \frac{1}{x}\]Trigonometric
Sine
Inserting sine into the basic formular results to as follows
\[f(x) = \sin x\] \[f'(x) = \lim_{\Delta x\rightarrow 0}\frac{\sin(x+\Delta x)-\sin(x)}{\Delta x}\]Using the Sum-to-product identity we get the following
\[f'(x) = \lim_{\Delta x\rightarrow 0}\frac{2\sin\left(\frac{x + \Delta x - x}{2}\right)\cos\left(\frac{x + \Delta x + x}{2}\right)}{\Delta x} = \lim_{\Delta x\rightarrow 0}\frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}} * \lim_{\Delta x\rightarrow 0}\cos\left(x + \frac{\Delta x}{2}\right)\]Using the Squeeze theorem we can get rid of sine and insert \(\Delta x \rightarrow 0\).
\[f'(x) = \cos(x)\]Cosine
Cosine can be expressed as follows
\[f(x) = \cos x = \sin\left(\frac{\pi}{2} - x\right)\]Deriving it creates the following
\[f'(x) = \frac{d}{dx}\sin\left(\frac{\pi}{2} - x\right) = \cos\left(\frac{\pi}{2} - x\right) * \frac{d}{dx}\left[\frac{\pi}{2} - x\right]\] \[f'(x) = \sin x * (-1) = -\sin x\]Inverse
The inverse of sine is expressed as follows
\[f(x) = \arcsin x\]If \(f(x) = \arcsin x\) then the following happens
\[\sin f(x) = x\] \[\frac{d}{dx}\sin f(x) = \frac{d}{dx}x = \cos f(x)f'(x) = 1\]Inserting the original equation lead to the following
\[1 = \cos\arcsin xf'(x)\] \[f'(x) = \frac{1}{\cos\arcsin x}\]Using a Pythagorean identity we can substitute for cosine
\[f'(x) = \frac{1}{\sqrt{1-\sin^2\arcsin x}} = \frac{1}{\sqrt{1-x^2}}\]