DIFFERENTIATION OF STANDARD FUNCTIONS

CBSE Class 11 - Mathematics - Limits and Derivatives Part-10

You have learnt theorems of differentiation part-9. Let us proceed further. Here is a list of differentiation of some standard functions:

$ \frac{\mathrm{d} }{\mathrm{d} x}(constant)=0 $

$ \frac{\mathrm{d} }{\mathrm{d} x}kx = k $ where k is constant

$ \frac{\mathrm{d} }{\mathrm{d} x}(x^{n})=nx^{n-1} $

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

$ \frac{\mathrm{d} }{\mathrm{d} x}(a^{x})=a^{x}\log _{e}a $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\log _{e}x)=\frac{1}{x} $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\log _{a}x)=\frac{1}{x \log_{e}a} $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin x \right )=\cos x $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos x \right )= -\sin x $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \tan x \right )= \sec ^{2}x $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cot x \right )= -cosec ^{2} x $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sec x \right )= \sec x\, \tan x $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \csc x \right )= -cosec x\, \cot x $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin^{-1}x \right )= \frac{1}{\sqrt{1-x^{2}}}\, ,-1< x< 1 $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos^{-1}x \right )= -\frac{1}{\sqrt{1-x^{2}}}\, ,-1< x< 1 $

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

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cot^{-1}x \right )= -\frac{1}{1+x^{2}} $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sec^{-1}x \right )= \frac{1}{\left | x \right |\sqrt{x^{2}-1}}\, ;\left | x \right |> 1 $

$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \csc^{-1}x \right )= \frac{-1}{\left | x \right |\sqrt{x^{2}-1}}\, ;\left | x \right |> 1 $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\sinh x)=\cosh x $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\cosh x)=\sinh x $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\tanh x)=\sec h^{2}x $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\coth x)= -\csc h^{2}x $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\sec h\, x)=\sec hx \, \tanh x $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\csc h x)= -\csc hx\, \coth x $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\sinh^{-1} x)= \frac{1}{\sqrt{1+x^{2}}} $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\cosh^{-1} x)= \frac{1}{\sqrt{x^{2}-1}} $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\tanh^{-1} x)= \frac{1}{1+x^{2}} $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\coth^{-1} x)= \frac{1}{x^{2}-1} $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\sec h^{-1} x)= -\frac{1}{x\sqrt{1-x^{2}}} $

$ \frac{\mathrm{d} }{\mathrm{d} x}(\csc h^{-1} x)= -\frac{1}{\left | x \right |\sqrt{x^{2}+1}} $

$ \frac{\mathrm{d} }{\mathrm{d} x}(e^{ax} \sin b\, x)= e^{ax}(a \sin bx + b \cos bx )=\sqrt{a^{2}+b^{2}}\, e^{ax}\sin (bx\, +\, \tan^{-1}b/a ) $

$ \frac{\mathrm{d} }{\mathrm{d} x}(e^{ax} \cos b\, x)= e^{ax}(a \cos bx - b \sin bx )=\sqrt{a^{2}+b^{2}}\, e^{ax}\cos (bx\, +\, \tan^{-1}b/a ) $

CBSE Papers, Questions, Answers, MCQ ...