You are here
Ex 5.2, 4 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Finding derivative of a function by chain rule
Chapter 5 Class 12 Continuity and Differentiability
Concept wise
- Checking continuity at a given point
- Checking continuity at any point
- Checking continuity using LHL and RHL
- Algebra of continous functions
- Continuity of composite functions
- Checking if funciton is differentiable
-
Finding derivative of a function by chain rule
- Finding derivative of Implicit functions
- Finding derivative of Inverse trigonometric functions
- Finding derivative of Exponential & logarithm functions
- Logarithmic Differentiation - Type 1
- Logarithmic Differentiation - Type 2
- Derivatives in parametric form
- Finding second order derivatives - Normal form
- Finding second order derivatives- Implicit form
- Proofs
- Verify Rolles theorem
- Verify Mean Value Theorem
Transcript
Ex 5.2, 4 Differentiate the functions with respect to x sec (tan ( βπ₯ )) Let π¦ = sec (tan βπ₯ ) We need to find Derivative of π¦ i.e. π¦β = (secβ‘γγ(tanγβ‘βπ₯)γ )^β² = γπ¬ππ γβ‘γ(πππ§β‘βπ)γ γπππ§ γβ‘γ(πππβ‘βπ)γ (tanβ‘βπ₯ )^β² = γsec γβ‘γ(tanβ‘βπ₯)γ γtan γβ‘γ(tanβ‘βπ₯)γ. ("sec2 " βπ₯ " . " (βπ₯)^β²) = γsec γβ‘γ(tanβ‘βπ₯)γ γtan γβ‘γ(tanβ‘βπ₯)γ. sec2 " " βπ₯ Γ 1/(2βπ₯) = (πππβ‘γ(πππβ‘βπ γ)πππβ‘γ(πππβ‘βπ γ)γπππγ^πβ‘βπ )/(πβπ)
