# Example 17 - Chapter 13 Class 11 Limits and Derivatives

Last updated at Nov. 30, 2019 by Teachoo

Last updated at Nov. 30, 2019 by Teachoo

Transcript

Example 17 Compute the derivative tan x. Let f(x) = tan x We need to find f’ (x) We know that f’(x) = lim┬(ℎ→0) f〖(𝑥 + ℎ) − f (x)〗/ℎ Here, f(x) = tan x f(x + ℎ) = tan (x + ℎ) Putting values f’ (x) = lim┬(ℎ→0) tan〖(𝑥 + ℎ) −tan𝑥 〗/ℎ = lim┬(ℎ→0) 1/ℎ ( tan (x + h) – tan x) = lim┬(ℎ→0) 1/ℎ (sin(𝑥 + ℎ)/cos(𝑥 + ℎ) − sin𝑥/cos𝑥 ) = lim┬(ℎ→0) 1/ℎ (〖cos x sin〗〖(𝑥 + ℎ) −〖 cos〗〖(𝑥 + ℎ) sin𝑥 〗 〗/cos〖(𝑥 + ℎ) cos𝑥 〗 ) = lim┬(ℎ→0) 1/ℎ (𝒔𝒊𝒏〖(𝒙 + 𝒉) 𝒄𝒐𝒔〖𝒙 − 𝒄𝒐𝒔〖(𝒙 + 𝒉). 〖 𝒔𝒊𝒏〗𝒙 〗 〗 〗/cos〖(𝑥 + ℎ) cos𝑥 〗 ) Using sin (A – B) = sin A cos B – cos B sin A Here A = x + h & B = x = lim┬(ℎ→0) 1/ℎ (𝐬𝐢𝐧(( 𝒙 + 𝒉 ) − 𝒙)/cos〖(𝑥 + ℎ) cos𝑥 〗 ) = lim┬(ℎ→0) 1/ℎ ((sin〖ℎ) 〗)/cos〖(𝑥 + ℎ) cos𝑥 〗 = lim┬(ℎ→0) sinℎ/ℎ 1/cos〖(𝑥 + ℎ) cos𝑥 〗 = (𝐥𝐢𝐦)┬(𝒉→𝟎) 𝒔𝒊𝒏𝒉/𝒉 " ×" lim┬(ℎ→0) 1/cos〖(𝑥 + ℎ) cos𝑥 〗 = 1 × lim┬(ℎ→0) 1/cos〖(𝑥 + ℎ) cos𝑥 〗 = lim┬(ℎ→0) 1/cos〖(𝑥 + ℎ) cos𝑥 〗 Putting ℎ = 0 = 1/cos〖(𝑥 + 0) cos𝑥 〗 = 1/〖cos x〗〖 .cos𝑥 〗 = 1/(〖𝑐𝑜𝑠〗^2 𝑥) = sec2x Hence , f’(x) = sec2x

Derivatives by formula - other trignometric

Chapter 13 Class 11 Limits and Derivatives

Concept wise

- Limits - Definition
- Limits - 0/0 form
- Limits - x^n formula
- Limits - Of Trignometric functions
- Limits - Limit exists
- Derivatives by 1st principle - At a point
- Derivatives by 1st principle - At a general point
- Derivatives by formula - x^n formula
- Derivatives by formula - sin & cos
- Derivatives by formula - other trignometric

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.