Examples

Chapter 13 Class 11 Limits and Derivatives
Serial order wise

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Example 19 Find the derivative of f from the first principle, where f is given by (i) f(x) = (2x + 3)/(x − 2) Let f (x) = (2x + 3)/(x − 2) We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = lim┬(h→0) f⁡〖(x + h) − f(x)〗/h Here, f (x) = (2x + 3)/(x − 2) So, f (x + h) = (2 (x + h) + 3)/((x + h)− 2) Putting values f’(x) = lim┬(h→0)⁡〖(((2 (𝑥 + ℎ)+3)/((𝑥 + ℎ)− 2)) − ((2𝑥 + 3)/(𝑥 − 2 )))/ℎ〗 = lim┬(h→0)⁡〖(((𝑥 − 2) (2 (𝑥 + ℎ) + 3)−(𝑥 + ℎ − 2) (2𝑥 + 3))/((𝑥 + ℎ −2) (𝑥 − 2)) )/ℎ〗 = lim┬(h→0)⁡〖((𝑥 −2)(2𝑥 +2ℎ + 3) − (𝑥 + ℎ −2)(2𝑥 +3))/(ℎ( 𝑥 + ℎ − 2 ) (𝑥 − 2))〗 = lim┬(h→0)⁡〖((𝑥 −2) ((2𝑥 + 3) + 2ℎ) − ((𝑥 −2)+ ℎ) (2𝑥 + 3) )/(ℎ( 𝑥 + ℎ − 2 ) (𝑥 − 2))〗 = lim┬(h→0)⁡〖((𝑥 − 2)(2𝑥 + 3) + (𝑥 −2)2ℎ− (𝑥 − 2) (2𝑥 + 3) − ℎ (2𝑥 + 3) )/(ℎ( 𝑥 + ℎ − 2 ) (𝑥 − 2))〗 = lim┬(h→0)⁡〖(2ℎ (𝑥 − 2) − ℎ (2𝑥 + 3) + (𝑥 − 2) (2𝑥 +3) − (𝑥 − 2) (2𝑥 + 3) )/(ℎ( 𝑥 + ℎ − 2 ) (𝑥 − 2))〗 = lim┬(h→0)⁡〖(h(2(x − 2)− (2x + 3)) + 0)/(ℎ( 𝑥 + ℎ − 2 ) (𝑥 − 2))〗 = 〖lim┬(h→0) 〗⁡〖(2(𝑥 − 2) − (2𝑥 + 3))/((𝑥 − 2) (𝑥 + ℎ − 2) )〗 = lim┬(h→0)⁡〖(2𝑥 − 4 − 2𝑥 − 3)/((𝑥 −2) (𝑥 + ℎ − 2) )〗 = lim┬(h→0)⁡〖(− 7)/((𝑥 −2) (𝑥 + ℎ − 2) )〗 Putting h = 0 = (− 7)/((𝑥 −2) (𝑥 + 0 − 2) ) = (− 7)/((𝑥 − 2) (𝑥 − 2)) = (− 𝟕)/(𝒙 − 𝟐)𝟐

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.