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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

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

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Concept wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo