Difference between revisions of "Derivative/Formulas"

Latest revision as of 15:23, 11 March 2022

$\frac d{dx}(cf(x)) = c\left(\frac d{dx} f(x)\right)$ where $c$ is a constant

$(f(x) + g(x))' = f'(x) + g'(x)$

$(f(x)-g(x))'=f'(x)-g'(x)$

$\left(u(x)\times v(x)\right)'=u(x)v'(x)+u'(x)v(x)$

$\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$(f(g(x)))' = f'(g(x))g'(x)$

$\frac d{dx} x^n = n x^{n-1}$

$\frac d{dx} (f(x))^n =n f(x)^{n-1} f'(x)$

$\frac d{dx} \sin x = \cos x$

$\frac d{dx} \cos x = -\sin x$

$\frac d{dx} \tan x = \sec^2 x$

$\frac d{dx} \sec x = \sec x \tan x$

$\frac d{dx} \csc x = -\csc x\cot x$

$\frac d{dx} \cot x = -\csc^2 x$

$\frac d{dx} e^x = e^x$

$\frac d{dx} a^x = (\ln a) a^x$

$\frac d{dx} \ln x = \frac 1x$

$\frac d{dx} \log_b x =\frac{\log_b e}{x}$

$\frac d{dx} \arcsin x = \frac 1{\sqrt{1-x^2}}$

$\frac d{dx} \arccos x = -\frac 1{\sqrt{1-x^2}}$

$\frac d{dx} \arctan x = \frac 1{1+x^2}$

$\frac d{dx} \mathrm{arcsec \ } x = \frac 1{\lvert x \rvert \sqrt{x^2-1}}$

$\frac d{dx} \mathrm{arccsc \ } x = - \frac 1{x\sqrt{x^2 - 1}}$

$\frac d{dx} \mathrm{arccot \ } x = - \frac 1{1+x^2}$

The following are commonly recognized notations for expressing the derivative of a function.

@@ Line 2: / Line 2: @@
 {| style="margin: 1em auto 1em auto; height:1000px"
-| <math>\frac d{dx}(cf(x)) = c\left(\frac d{dx} f(x)\right)</math> where c is a constant
+| <math>\frac d{dx}(cf(x)) = c\left(\frac d{dx} f(x)\right)</math> where <math>c</math> is a constant
 |-
 | <math>(f(x) + g(x))' = f'(x) + g'(x)</math>
@@ Line 44: / Line 44: @@
 | <math>\frac d{dx} \arctan x = \frac 1{1+x^2}</math>
 |-
-| <math>\frac d{dx} \mathrm{arcsec \ } x = \frac 1{\mid x \mid\sqrt{x^2-1}}</math>
+| <math>\frac d{dx} \mathrm{arcsec \ } x = \frac 1{\lvert x \rvert \sqrt{x^2-1}}</math>
 |-
 | <math>\frac d{dx} \mathrm{arccsc \ } x = - \frac 1{x\sqrt{x^2 - 1}}</math>