Formulas of Differential Calculus | Derivative Rules | ক্যালকুলাস এর সূত্রসমূহ

Formulas of Differential Calculus | Derivative Rules | ক্যালকুলাস এর সূত্রসমূহ

$\frac{d}{dx}(c)$  = 0                             [where, c is single constant]

$\frac{d}{dx}(cu)$ = c $\frac {d}{dx}$ (u)   [where, c is with a variable]

$\frac{d}{dx}$ (xⁿ) = nx^n-1

$\frac{d}{dx}$ $(\sqrt {x})= \frac {1}{2\sqrt{x}}$

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

$\frac{d}{dx}$ (e^mx)= me^mx

$\frac{d}{dx}$ (e^-mx)= -me^-mx

$\frac{d}{dx}$ (log_a x)= $\frac {1}{x}$ log_ae

$\frac{d}{dx}$ (log x)= $\frac {1}{x}$

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

$\frac{d}{dx}$ (sin mx)= m cos mx

$\frac{d}{dx}$ (cos x)= - sin x

$\frac{d}{dx}$ (cos mx)= - m sin mx

$\frac{d}{dx}$ (sec x)= sec x. tan x

$\frac{d}{dx}$ (cosec x)= -cosec x. cot x

$\frac{d}{dx}$ (tan x)= sec² x

$\frac{d}{dx}3$ (cot x)= -cosec² x

$\frac{d}{dx}$ (sin^-1) = $1  \over { \sqrt{1-x^2 }}$

$\frac{d}{dx}$ (cos^-1)= $-1  \over { \sqrt{1-x^2 }}$

$\frac{d}{dx}$ (tan^-1)= $\frac{1}{1+x^2}$

$\frac{d}{dx}$ (uv)= u $\frac {d}{dx}$ (v) + v $\frac {d}{dx}$ (u)

$\frac{d}{dx}$ (uvw) = vw $\frac {d}{dx}$ (u) + uw $\frac {d}{dx}$ (v) + uv $\frac {d}{dx}$( w )

$\frac {d}{dx}$ ($\frac {u}{v})$ = $\frac{v\frac{\mathrm{d}}{\mathrm{d} x} ( u )-u\frac{\mathrm{d} }{\mathrm{d} x} ( v )}{v^{2}}$


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