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Prove that Derivative of tan x is sec^2 x - by First Principle

Example 17 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Example 17 - Chapter 13 Class 11 Limits and Derivatives - Part 3
Example 17 - Chapter 13 Class 11 Limits and Derivatives - Part 4

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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) −tan⁡𝑥 〗/ℎ = lim┬(ℎ→0) 1/ℎ ( tan (x + h) – tan x) = lim┬(ℎ→0) 1/ℎ (sin⁡(𝑥 + ℎ)/cos⁡(𝑥 + ℎ) − sin⁡𝑥/cos⁡𝑥 ) = lim┬(ℎ→0) 1/ℎ (〖cos x sin〗⁡〖(𝑥 + ℎ) −〖 cos〗⁡〖(𝑥 + ℎ) sin⁡𝑥 〗 〗/cos⁡〖(𝑥 + ℎ) cos⁡𝑥 〗 ) = lim┬(ℎ→0) 1/ℎ (𝒔𝒊𝒏⁡〖(𝒙 + 𝒉) 𝒄𝒐𝒔⁡〖𝒙 − 𝒄𝒐𝒔⁡〖(𝒙 + 𝒉). 〖 𝒔𝒊𝒏〗⁡𝒙 〗 〗 〗/cos⁡〖(𝑥 + ℎ) cos⁡𝑥 〗 ) 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) cos⁡𝑥 〗 ) = lim┬(ℎ→0) 1/ℎ ((sin⁡〖ℎ) 〗)/cos⁡〖(𝑥 + ℎ) cos⁡𝑥 〗 = lim┬(ℎ→0) sin⁡ℎ/ℎ 1/cos⁡〖(𝑥 + ℎ) cos⁡𝑥 〗 = (𝐥𝐢𝐦)┬(𝒉→𝟎) 𝒔𝒊𝒏⁡𝒉/𝒉 " ×" lim┬(ℎ→0) 1/cos⁡〖(𝑥 + ℎ) cos⁡𝑥 〗 = 1 × lim┬(ℎ→0) 1/cos⁡〖(𝑥 + ℎ) cos⁡𝑥 〗 = lim┬(ℎ→0) 1/cos⁡〖(𝑥 + ℎ) cos⁡𝑥 〗 Putting ℎ = 0 = 1/cos⁡〖(𝑥 + 0) cos⁡𝑥 〗 = 1/〖cos x〗⁡〖 .cos⁡𝑥 〗 = 1/(〖𝑐𝑜𝑠〗^2 𝑥) = sec2x Hence , f’(x) = sec2x

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.