Logarithmic Differentiation - Type 1

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

### Transcript

Example 40 (Method 1) Differentiate the following π€.π.π‘. π₯. (i) cos^(β1) (sinβ‘π₯) Let π(π₯) = cos^(β1) (sinβ‘π₯) π(π₯) = cos^(β1) (γcos γβ‘(π/2 βπ₯) ) π(π) = π/π βπ Differentiating π€.π.π‘.π₯ πβ(π₯) = (π (π/2))/ππ₯ β (π(π₯))/ππ₯ πβ(π₯) = 0 β 1 πβ(π) = β 1(π΄π  γ π ππ π γβ‘γ=γπππ  γβ‘γ(π/2 βπ₯)γ γ ) ("As " (π(π₯))/ππ₯ " = 1 & " π/2 " is constant" ) Example 40 (Method 2) Differentiate the following π€.π.π‘. π₯. (i) cos^(β1) (sinβ‘π₯) Let π(π₯) = cos^(β1) (sinβ‘π₯) Differentiating π€.π.π‘.π₯ πβ²(π₯) = (β1)/β(1 β γ(sinβ‘π₯)γ^2 ) Γ (sinβ‘π₯ )^β² πβ²(π₯) = (β1)/β(1 β sin^2β‘π₯ ) Γcosβ‘π₯ πβ²(π₯) = (β1)/β(cos^2β‘π₯ ) Γcosβ‘π₯ πβ²(π₯) = (β1)/cosβ‘π₯ Γcosβ‘π₯ πβ(π) = β1

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#### 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.