Ex 12.2, 10 - Find derivative of cos x from first principle - Teachoo - Ex 12.2

part 2 - Ex 12.2, 10 - Ex 12.2 - Serial order wise - Chapter 12 Class 11 Limits and Derivatives
part 3 - Ex 12.2, 10 - Ex 12.2 - Serial order wise - Chapter 12 Class 11 Limits and Derivatives

Remove Ads Take short quiz All Quiz and Worksheets
Teachoo Β· Class 11 Explore Class 11

Transcript

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) = (π‘™π‘–π‘š)┬(β„Žβ†’0) 𝑓⁑〖(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)γ€—/β„Ž Here, f (x) = cos x So, f (x + h) = cos (x + h) Putting values, f’ (x) = lim┬(hβ†’0)⁑〖(𝒄𝒐𝒔 (𝒙 + 𝒉) βˆ’γ€– 𝒄𝒐𝒔〗⁑𝒙)/hγ€— Using cos A – cos B = – 2 sin ((𝐴 + 𝐡)/2) sin ((𝐴 βˆ’ 𝐡)/2) = lim┬(hβ†’0)⁑〖(βˆ’πŸ π’”π’Šπ’((𝒙 + (𝒙 + 𝒉))/𝟐) . π’”π’Šπ’(((𝒙 + 𝒉) βˆ’ 𝒙)/𝟐))/hγ€— = lim┬(hβ†’0)⁑〖(βˆ’2 𝑠𝑖𝑛((2π‘₯ + β„Ž)/2) . 𝑠𝑖𝑛(β„Ž/2))/hγ€— = lim┬(hβ†’0)β‘γ€–βˆ’2 sin⁑((2π‘₯ + β„Ž)/2).γ€–sin γ€—β‘γ€–β„Ž/2γ€—/β„Žγ€— = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2).γ€–sin γ€—β‘γ€–β„Ž/2γ€—/(β„Ž/2)γ€— Using (π‘™π‘–π‘š)┬(π‘₯β†’0)⁑〖 𝑠𝑖𝑛⁑π‘₯/π‘₯γ€—=1 Replacing x by β„Ž/2 β‡’ (π‘™π‘–π‘š)┬(β„Žβ†’0) 𝑠𝑖𝑛⁑〖 β„Ž/2γ€—/(( β„Ž)/2) = 1 = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2).(π₯𝐒𝐦)┬(π‘β†’πŸŽ) 〖𝐬𝐒𝐧 〗⁑〖𝒉/πŸγ€—/(𝒉/𝟐)γ€— = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2).πŸγ€— = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2) γ€— Putting h = 0 = βˆ’sin⁑((2π‘₯ +0)/2) = βˆ’sin⁑(2π‘₯/2) = – sin x ∴ f’(x) = –sin x

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.