The derivative of any polynomial function of one variable is easily obtained. If c\in\R (or a constant function) and f,g:D\to\R are both differentiable on some set D' , then so are c\cdot f , f+g , -f , and f\cdot g . If, in addition, g is nonzero on D' , then \frac{1}{g} (and also \frac{f}{g}) are differentiable on D' . Also, if f is differentiable on g\left(D'\right), then f\circ g is differentiable on D' . For the trivial case of f(x)=a , for some constant a (a degree 0 polynomial):

a'=0 [Proof]

For any r\in\R :

(x^r)'=rx^{r-1} [Proof]

Which covers any single variable polynomial function. Derivatives of non-polynomial functions require additional rules.

For any real-valued differentiable functions f(x),g(x) :

  • (-f(x))'=-f'(x)
  • \left(\frac{1}{g(x)}\right)'=-\frac{g'(x)}{g(x)^2}

Trigonometric functions:

  • \frac{d}{dx}(\cos(x))=-\sin(x)
  • \frac{d}{dx}(\tan(x))=\sec^2(x)
  • \frac{d}{dx}(\csc(x))=-\csc(x)\cot(x)
  • \frac{d}{dx}(\sec(x))=\sec(x)\tan(x)
  • \frac{d}{dx}(\cot(x))=-\csc^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}(\arcsec(x))=\frac{1}{|x|\sqrt{x^2-1}}
  • \frac{d}{dx}(\arccsc(x))=-\frac{1}{|x|\sqrt{x^2-1}}
  • \frac{d}{dx}(\arccot(x))=-\frac{1}{1+x^2}

Logarithmic and exponential functions:

  • \frac{d}{dx}(e^x)=e^x
  • \frac{d}{dx}(a^x)=a^x\ln(a)
  • \frac{d}{dx}(\log_a(x))=\frac{1}{\ln(a)x}