Derivatives

The derivative at point $x$ is given by $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Differentiation formulas

$\frac{d}{dt}(k)=0,\text{for any constant }k$ $\frac{d}{dt}(x^n)=n{x}^{x-1},\text{for any constant }n$ $\frac{d}{dx}(kf(x))=k\frac{df}{dx},\text{for any constant }k$ $\frac{d}{dx}[f(x)+g(x)]=\frac{df}{dx}+\frac{dg}{dx}\text{(assuming that these both exist)}$ $\frac{d}{dx}(\sin x)=\cos x$ $\frac{d}{dx}(\cos x)=-\sin x$ $\frac{d}{dx}(e^x)=e^x$ $\frac{d}{dx}(\tan x)={\sec }^{2}x$ $\frac{d}{dx}(\cot x)=-{\csc }^{2}x$ $\frac{d}{dx}(\sec x)=\sec x\tan x$ $\frac{d}{dx}(\csc x)=-\csc x\cot x$ $\frac{d}{dx}(\cot x)=-{\csc }^{2}x$ $\frac{d}{dx}(\cosh x)=\sinh x$ $\frac{d}{dx}(\sinh x)=\cosh x$ $\frac{d}{dx}(\tanh x)=sechx$ $\frac{d}{dx}(\ln x)=\frac{1}{x}$ $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$ $\frac{d}{dx}({\cos }^{-1}x)=-\frac{1}{\sqrt{1-x^2}}$ $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$ $\frac{d}{dx}(a^x)=(\ln a)a^x$ $\frac{d}{dx}({\log }_{a}x)=\frac{1}{(\ln a)x}$

The product Rule

$\frac{d}{dx}[f(x)g(x)]=\frac{df}{dx}\cdot g+f\cdot \frac{dg}{dx}$

The Quotient Rule

$\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{1}{[g(x){]}^2}[\frac{df}{dx}\cdot g-f\cdot \frac{dg}{dx}]$

Chain Rule

Implicit Differentiation

For functions which are defined implicitly, such as $x^2+y^2=4$, we can differentiate both sides of the equation with respect to $x$, applying the Chain Rule whenever necessary.

Logarithmic Differentiation

For some cases, such as $(\cos x{)}^x$, we cannot differentiate directly. We can do $y=(\cos x{)}^x$ $\ln y=x\ln (\cos x)$ $\frac{1}{y}\frac{dy}{dx}=\dots$

Derivatives of Inverses

$\frac{dy}{dx}=\frac{1}{dx/dy}$ However, this result is only true for first-order derivatives. It doesn’t work for higher order derivatives.

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(\frac{1}{dx/dy})$

Related

• Function
• Curve Sketching
• Differential Calculus Theorems
• Partial Derivatives
• Directional Derivatives