Example 17 - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

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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