Example 20 (i) - Chapter 13 Class 11 Limits and Derivatives

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Example 20 Find the derivative of f(x) from the first principle, where f(x) is (i) sin x + cos x Given f (x) = sin x + cos x We need to find Derivative of f(x) We know that f’(x) = lim┬(h→0) 𝑓⁡〖(𝑥 + ℎ) − 𝑓(𝑥)〗/ℎ Here, f (x) = sin x + cos x f (x + h) = sin (x + h) + cos (x + h) Putting values f’(x) = lim┬(h→0)⁡〖(sin⁡〖(𝑥 + ℎ)〗 + cos⁡(𝑥 + ℎ) − (sin⁡𝑥 + cos⁡〖𝑥)〗)/ℎ〗 Using sin (A + B) = sin A cos B + cos A sin B & cos (A + B) = cos A cos B – sin A sin B = lim┬(h→0)⁡〖sin⁡〖𝑥 cos⁡〖ℎ +〖 cos〗⁡〖𝑥 sin⁡〖ℎ + cos⁡〖𝑥 cos⁡〖ℎ − sin⁡〖𝑥 〖 sin〗⁡〖ℎ −〖 sin〗⁡〖𝑥 −〖 cos〗⁡𝑥 〗 〗 〗 〗 〗 〗 〗 〗 〗/h〗 = lim┬(h→0)⁡〖cos⁡〖𝑥 sin⁡〖ℎ −〖 sin〗⁡〖𝑥 sin⁡〖ℎ + sin⁡〖𝑥 cos⁡〖ℎ − sin⁡〖𝑥 +〖 cos〗⁡〖𝑥 cos⁡〖ℎ −〖 cos〗⁡𝑥 〗 〗 〗 〗 〗 〗 〗 〗 〗/h〗 = lim┬(h→0)⁡〖sin⁡〖ℎ 〖(cos〗⁡〖𝑥 − sin⁡〖𝑥) + sin⁡〖𝑥 (cos⁡〖ℎ − 1) + cos⁡𝑥 (cos⁡〖ℎ − 1)〗 〗 〗 〗 〗 〗/h〗 = lim┬(h→0)⁡(sin⁡〖ℎ 〖(cos〗⁡〖𝑥 − sin⁡〖𝑥)〗 〗 〗/h+sin⁡〖𝑥 (cos⁡〖ℎ − 1)〗 〗/h+cos⁡〖𝑥 (cos⁡〖ℎ − 1)〗 〗/ℎ)" " = lim┬(h→0)⁡〖sin⁡〖ℎ 〖(cos〗⁡〖𝑥 −〖 sin〗⁡〖𝑥)〗 〗 〗/h+lim┬(h→0) sin⁡〖𝑥 (cos⁡〖ℎ − 1)〗 〗/h+lim┬(h→0) cos⁡〖𝑥 (cos⁡〖ℎ − 1)〗 〗/ℎ〗 = lim┬(h→0)⁡〖"(cos x – sin x)" sin⁡ℎ/ℎ+lim┬(h→0) "(– sin x)" ((1 − cos⁡〖ℎ)〗)/ℎ〗+lim┬(h→0) "(– cos x)" ((1 − cos⁡〖ℎ)〗)/ℎ = "(cos x – sin x)" (𝐥𝐢𝐦)┬(𝐡→𝟎)⁡〖𝐬𝐢𝐧⁡𝒉/𝒉−sin⁡〖𝑥 (𝐥𝐢𝐦)┬(𝐡→𝟎) 〗 ((𝟏 − 𝒄𝒐𝒔⁡〖𝒉)〗)/𝒉−cos⁡𝑥 (𝐥𝐢𝐦)┬(𝐡→𝟎) ((𝟏 − 𝒄𝒐𝒔⁡〖𝒉)〗)/𝒉〗Using (𝑙𝑖𝑚)┬(ℎ→0) 𝑠𝑖𝑛⁡ℎ/ℎ = 1 & (𝑙𝑖𝑚)┬(ℎ→0) 〖(1 − 𝑐𝑜𝑠〗⁡〖ℎ)〗/ℎ = 0 Using (𝑙𝑖𝑚)┬(ℎ→0) 𝑠𝑖𝑛⁡ℎ/ℎ = 1 & (𝑙𝑖𝑚)┬(ℎ→0) 〖(1 − 𝑐𝑜𝑠〗⁡〖ℎ)〗/ℎ = 0