Differentiation

To differentiate something means to see how fast it changes with respect to (w.r.t) something else changing, for example, time $t$ . Imagine you step forward in time by a very small step $h$ . As $h\rightarrow 0$ , the change in $x$ becomes the derivative of $x$ with respect to time.

\frac{d}{dt}f(t)=\lim_{h\to 0} \frac{f(t+h)-f(t)}{h}

To do this, us humans developed some nifty tricks. Here’s all of them.

\begin{align*} \frac{d}{dx}[f(x) \pm g(x)] & = f'(x) \pm g'(x) \\ \frac{d}{dx}[x^n] & = nx^{n-1} \\ \frac{d}{dx}[f(x)g(x)] & = f'(x)g(x) + f(x)g'(x) \\ \frac{d}{dx}[f(g(x))] & = f'(g(x))\,g'(x) \\ \frac{d}{dx}[e^x] & = e^x \\ \frac{d}{dx}[a^x] & = a^x \ln(a) \\ \frac{d}{dx}[\ln(x)] & = \frac{1}{x} \\ \frac{d}{dx}[\log_a(x)] & = \frac{1}{x \ln(a)} \\ \frac{d}{dx}[\sin(x)] & = \cos(x) \\ \frac{d}{dx}[\cos(x)] & = -\sin(x) \\ \frac{d}{dx}[\tan(x)] & = \left(\frac{1}{\cos(x)}\right)^2 \\ \frac{d}{dx}\left(\frac{1}{\tan(x)}\right) & = \frac{1}{\sin^2 (x)} \\ \frac{d}{dx}\left(\frac{1}{\cos(x)}\right) & = \frac{\sin x}{\cos^2 x} \\ \frac{d}{dx}\left(\frac{1}{\sin(x)}\right) & = -\frac{\cos x}{\sin^2 x} \\ \frac{d}{dx}[\arcsin(x)] & = \frac{1}{\sqrt{1 - x^2}} \\ \frac{d}{dx}[\arccos(x)] & = -\frac{1}{\sqrt{1 - x^2}} \\ \frac{d}{dx}[\arctan(x)] & = \frac{1}{1 + x^2} \\ \frac{d}{dx}[\operatorname{arccot}(x)] & = -\frac{1}{1 + x^2} \\ \frac{d}{dx}[\operatorname{arcsec}(x)] & = \frac{1}{|x|\sqrt{x^2 - 1}} \\ \frac{d}{dx}[\operatorname{arccsc}(x)] & = -\frac{1}{|x|\sqrt{x^2 - 1}} \\ \frac{d}{dx}[f(x)^{g(x)}] & = f(x)^{g(x)} \left[g'(x)\ln(f(x)) + \frac{g(x)\,f'(x)}{f(x)}\right] \\ \frac{d}{dx}[f^{-1}(x)] & = \frac{1}{f'(f^{-1}(x))} \\ \frac{d^n}{dx^n}[f(x)g(x)] & = \sum_{k=0}^{n} \binom{n}{k} f^{(k)}(x)\,g^{(n-k)}(x) \end{align*}