Differentiation Formulas
Last updated at April 5, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Differentiation forms the basis of calculus, and we need its formulas to solve problems.

We have prepared a list of all the Formulas

Basic Differentiation
Formulas

Differentiation of Log and Exponential Function

Differentiation of Trigonometry Functions

Differentiation of Inverse Trigonometry Functions

Differentiation Rules

Transcript

ππ/ππ₯=0
(π(π₯))/ππ₯=1
(π(ππ₯))/ππ₯=π
(π(π₯^π))/ππ₯=ππ₯^(π β 1)
(π(π^π₯))/ππ₯=π^π₯
(π(lnβ‘γ(π₯)γ))/ππ₯=1/π₯
(π(π^π₯))/ππ₯=π^π₯ γ logγβ‘π
(π(π₯^π₯))/ππ₯=π₯^π₯ (1+lnβ‘π₯)
(π(log_πβ‘π₯))/ππ₯=1/π₯Γ1/lnβ‘π
(π(sinβ‘π₯))/ππ₯=cosβ‘π₯
(π(cosβ‘π₯))/ππ₯=sinβ‘π₯
(π(tanβ‘π₯))/ππ₯=sec^2β‘π₯
(π(cotβ‘π₯))/ππ₯=βcosec^2β‘π₯ " "
(π(secβ‘π₯))/ππ₯=secβ‘π₯ tanβ‘π₯
(π(cosecβ‘π₯))/ππ₯=γβcosecγβ‘π₯ cotβ‘π₯
(π(sin^(β1)β‘π₯))/ππ₯= 1/β(1 β π₯^2 )
(π(cos^(β1)β‘π₯))/ππ₯= (β1)/β(1 β π₯^2 )
(π(tan^(β1)β‘π₯))/ππ₯= 1/(1 + π₯^2 )
(π(cot^(β1)β‘π₯))/ππ₯= (β1)/(1 +γ π₯γ^2 )
(π(sec^(β1)β‘π₯))/ππ₯= 1/(|π₯| β(π₯^2 β 1))
(π(cosec^(β1)β‘π₯))/ππ₯= (β1)/(π₯β(π₯^2 β 1))
Product Rule
π/ππ₯(π(π₯) π(π₯))=π^β² (π₯) π(π₯)+π(π₯) π^β² (π₯)
Quotient Rule
π/ππ₯ (π(π₯)/π(π₯) )=(π^β² (π₯) π(π₯) β π(π₯) π^β² (π₯))/(π (π₯))^2
Chain Rule
(π(π(π(π₯))))/ππ₯=π^β² (π(π₯)) π^β² (π₯)
First Derivative Rule
fβ(x) = (πππ)β¬(ββ0) πβ‘γ(π₯ + β) β π(π₯)γ/β

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