Examples

Chapter 12 Class 11 Limits and Derivatives
Serial order wise

### 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