Finding derivative of a function by chain rule

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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

We have prepared a list of all the Formulas

## Differentiation Rules

### Transcript

ππ/ππ₯=0 (π(π₯))/ππ₯=1 (π(ππ₯))/ππ₯=π (π(π₯^π))/ππ₯=ππ₯^(π β 1) (π(π^π₯))/ππ₯=π^π₯ (π(lnβ‘γ(π₯)γ))/ππ₯=1/π₯ (π(π^π₯))/ππ₯=π^π₯ γ logγβ‘π (π(π₯^π₯))/ππ₯=π₯^π₯ (1+lnβ‘π₯) (π(log_πβ‘π₯))/ππ₯=1/π₯Γ1/lnβ‘π (π(sinβ‘π₯))/ππ₯=cosβ‘π₯ (π(cosβ‘π₯))/ππ₯=sinβ‘π₯ (π(tanβ‘π₯))/ππ₯=sec^2β‘π₯ (π(cotβ‘π₯))/ππ₯=βcosec^2β‘π₯ " " (π(secβ‘π₯))/ππ₯=secβ‘π₯ tanβ‘π₯ (π(cosecβ‘π₯))/ππ₯=γβcosecγβ‘π₯ cotβ‘π₯ (π(sin^(β1)β‘π₯))/ππ₯= 1/β(1 β π₯^2 ) (π(cos^(β1)β‘π₯))/ππ₯= (β1)/β(1 β π₯^2 ) (π(tan^(β1)β‘π₯))/ππ₯= 1/(1 + π₯^2 ) (π(cot^(β1)β‘π₯))/ππ₯= (β1)/(1 +γ π₯γ^2 ) (π(sec^(β1)β‘π₯))/ππ₯= 1/(|π₯| β(π₯^2 β 1)) (π(cosec^(β1)β‘π₯))/ππ₯= (β1)/(π₯β(π₯^2 β 1)) Product Rule π/ππ₯(π(π₯) π(π₯))=π^β² (π₯) π(π₯)+π(π₯) π^β² (π₯) Quotient Rule π/ππ₯ (π(π₯)/π(π₯) )=(π^β² (π₯) π(π₯) β π(π₯) π^β² (π₯))/(π (π₯))^2 Chain Rule (π(π(π(π₯))))/ππ₯=π^β² (π(π₯)) π^β² (π₯) First Derivative Rule fβ(x) = (πππ)β¬(ββ0) πβ‘γ(π₯ + β) β π(π₯)γ/β