Check sibling questions

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

Basic Differentiation Formulas.jpg

Differentiation of Log and Exponential Function

Differentiation Formulas - Part 2

Differentiation of Trigonometry Functions

Differentiation Formulas - Part 3

Differentiation of Inverse Trigonometry Functions

Differentiation Formulas - Part 4

Differentiation Rules

Differentiation Formulas - Part 5

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


π‘‘π‘˜/𝑑π‘₯=0 (𝑑(π‘₯))/𝑑π‘₯=1 (𝑑(π‘˜π‘₯))/𝑑π‘₯=π‘˜ (𝑑(π‘₯^𝑛))/𝑑π‘₯=𝑛π‘₯^(𝑛 βˆ’ 1) (𝑑(𝑒^π‘₯))/𝑑π‘₯=𝑒^π‘₯ (𝑑(ln⁑〖(π‘₯)γ€—))/𝑑π‘₯=1/π‘₯ (𝑑(π‘Ž^π‘₯))/𝑑π‘₯=π‘Ž^π‘₯ γ€– logγ€—β‘π‘Ž (𝑑(π‘₯^π‘₯))/𝑑π‘₯=π‘₯^π‘₯ (1+ln⁑π‘₯) (𝑑(log_π‘Žβ‘π‘₯))/𝑑π‘₯=1/π‘₯Γ—1/lnβ‘π‘Ž (𝑑(sin⁑π‘₯))/𝑑π‘₯=cos⁑π‘₯ (𝑑(cos⁑π‘₯))/𝑑π‘₯=sin⁑π‘₯ (𝑑(tan⁑π‘₯))/𝑑π‘₯=sec^2⁑π‘₯ (𝑑(cot⁑π‘₯))/𝑑π‘₯=βˆ’cosec^2⁑π‘₯ " " (𝑑(sec⁑π‘₯))/𝑑π‘₯=sec⁑π‘₯ tan⁑π‘₯ (𝑑(cosec⁑π‘₯))/𝑑π‘₯=γ€–βˆ’cosec〗⁑π‘₯ cot⁑π‘₯ (𝑑(sin^(βˆ’1)⁑π‘₯))/𝑑π‘₯= 1/√(1 βˆ’ π‘₯^2 ) (𝑑(cos^(βˆ’1)⁑π‘₯))/𝑑π‘₯= (βˆ’1)/√(1 βˆ’ π‘₯^2 ) (𝑑(tan^(βˆ’1)⁑π‘₯))/𝑑π‘₯= 1/(1 + π‘₯^2 ) (𝑑(cot^(βˆ’1)⁑π‘₯))/𝑑π‘₯= (βˆ’1)/(1 +γ€– π‘₯γ€—^2 ) (𝑑(sec^(βˆ’1)⁑π‘₯))/𝑑π‘₯= 1/(|π‘₯| √(π‘₯^2 βˆ’ 1)) (𝑑(cosec^(βˆ’1)⁑π‘₯))/𝑑π‘₯= (βˆ’1)/(π‘₯√(π‘₯^2 βˆ’ 1)) Product Rule 𝑑/𝑑π‘₯(𝑓(π‘₯) 𝑔(π‘₯))=𝑓^β€² (π‘₯) 𝑔(π‘₯)+𝑓(π‘₯) 𝑔^β€² (π‘₯) Quotient Rule 𝑑/𝑑π‘₯ (𝑓(π‘₯)/𝑔(π‘₯) )=(𝑓^β€² (π‘₯) 𝑔(π‘₯) βˆ’ 𝑓(π‘₯) 𝑔^β€² (π‘₯))/(𝑔 (π‘₯))^2 Chain Rule (𝑑(𝑓(𝑔(π‘₯))))/𝑑π‘₯=𝑓^β€² (𝑔(π‘₯)) 𝑔^β€² (π‘₯) First Derivative Rule f’(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) 𝑓⁑〖(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)γ€—/β„Ž

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.