y=(cosx)^2 differentiated, chain rule version
y=(cosx)^2 differentiated, chain rule version

Ex 12.2

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Ex 12.2, 4 (i) Important

Ex 12.2, 4 (ii)

Ex 12.2, 4 (iii) Important

Ex 12.2, 4 (iv)

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Ex 12.2, 7 (i) Important

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Ex 12.2, 9 (i)

Ex 12.2, 9 (ii) Important

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Ex 12.2, 10 Important You are here

Ex 12.2, 11 (i)

Ex 12.2, 11 (ii) Important

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Ex 12.2

Last updated at May 29, 2023 by Teachoo

Ex 12.2, 10 Find the derivative of cos x from first principle. Let f (x) = cos x We need to find f’(x) We know that f’(x) = (𝑙𝑖𝑚)┬(ℎ→0) 𝑓〖(𝑥 + ℎ) − 𝑓(𝑥)〗/ℎ Here, f (x) = cos x So, f (x + h) = cos (x + h) Putting values, f’ (x) = lim┬(h→0)〖(𝒄𝒐𝒔 (𝒙 + 𝒉) −〖 𝒄𝒐𝒔〗𝒙)/h〗 Using cos A – cos B = – 2 sin ((𝐴 + 𝐵)/2) sin ((𝐴 − 𝐵)/2) = lim┬(h→0)〖(−𝟐 𝒔𝒊𝒏((𝒙 + (𝒙 + 𝒉))/𝟐) . 𝒔𝒊𝒏(((𝒙 + 𝒉) − 𝒙)/𝟐))/h〗 = lim┬(h→0)〖(−2 𝑠𝑖𝑛((2𝑥 + ℎ)/2) . 𝑠𝑖𝑛(ℎ/2))/h〗 = lim┬(h→0)〖−2 sin((2𝑥 + ℎ)/2).〖sin 〗〖ℎ/2〗/ℎ〗 = lim┬(h→0)〖−sin((2𝑥 + ℎ)/2).〖sin 〗〖ℎ/2〗/(ℎ/2)〗 Using (𝑙𝑖𝑚)┬(𝑥→0)〖 𝑠𝑖𝑛𝑥/𝑥〗=1 Replacing x by ℎ/2 ⇒ (𝑙𝑖𝑚)┬(ℎ→0) 𝑠𝑖𝑛〖 ℎ/2〗/(( ℎ)/2) = 1 = lim┬(h→0)〖−sin((2𝑥 + ℎ)/2).(𝐥𝐢𝐦)┬(𝐡→𝟎) 〖𝐬𝐢𝐧 〗〖𝒉/𝟐〗/(𝒉/𝟐)〗 = lim┬(h→0)〖−sin((2𝑥 + ℎ)/2).𝟏〗 = lim┬(h→0)〖−sin((2𝑥 + ℎ)/2) 〗 Putting h = 0 = −sin((2𝑥 +0)/2) = −sin(2𝑥/2) = – sin x ∴ f’(x) = –sin x

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