Basic Differentiation Rules

$$\frac{d}{dx}\left[cu\right]=cu' \frac{d}{dx}\left[u\pm v\right]=u'\pm v'$$

$$\frac{d}{dx}\left[uv\right]=uv'+vu' \frac{d}{dx}\left[\frac{u}{v}\right]=\frac{vu'-uv'}{v^2}$$

$$\frac{d}{dx}\left[c\right]=0 \frac{d}{dx}\left[u^n\right]=nu^{n-1}u'$$

$$\frac{d}{dx}\left[x\right]=1 \frac{d}{dx}\left[\left\lvert u \right\rvert \right]=\frac{u}{\left\lvert u \right\rvert }\left(u'\right),\quad u\ne 0$$

$$\frac{d}{dx}\left[\ln u\right]=\frac{u'}{u} \frac{d}{dx}\left[e^u\right]=e^uu'$$

$$\frac{d}{dx}\left[\sin u\right]=\left(\cos u\right)u' \frac{d}{dx}\left[\cos u\right]= -\!\left(\sin u\right)u'$$

$$\frac{d}{dx}\left[\tan u\right]=\left(\sec^2u\right)u' \frac{d}{dx}\left[\cot u\right]= -\!\left(\csc^2u\right)u'$$

$$\frac{d}{dx}\left[\sec u\right]=\left(\sec u \tan u\right)u' \frac{d}{dx}\left[\csc u\right]= -\!\left(\csc u \cot u\right)u'$$

$$\frac{d}{dx}\left[\arcsin u\right]=\frac{u'}{\sqrt{1-u^2}} \frac{d}{dx}\left[\arccos u\right]=\frac{-u'}{\sqrt{1-u^2}}$$

$$\frac{d}{dx}\left[{\mathrm{arccsc}}\frac{u^2-1}{\sqrt{1-u^2}}\right]=\frac{u'}{\left\lvert u \right\rvert \sqrt{u^2-1}} \frac{d}{dx}\left[{\mathrm{arccsc}}u\right]=\frac{-u'}{\left\lvert u \right\rvert \sqrt{u^2-1}}$$

Chain Rule

If $y=f(u)$ is a differentiable function of $u$ and $u=g(x)$ is a differentiable function of $x$ , then $y=f(g(x))$ is a differentiable function of $x$ and

$$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}$$

or equivalently,

$$\frac{d}{dx}\left[f(g(x))\right]=f'(g(x)) g'(x).$$