Misc 1 (iii) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Transcript

Misc 1 Find the derivative of the following functions from first principle: (iii) sin (x + 1) Let f (x) = sin (x + 1) 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 + 1) So, f (x + h) = sin ((x + h) + 1) Putting values f’(x) = lim┬(h→0)⁡〖sin⁡〖(( 𝑥 + ℎ ) + 1 ) −〖 sin〗⁡〖 (𝑥 + 1)〗 〗/h〗 = lim┬(h→0)⁡〖(𝑠𝑖𝑛( 𝑥 + 1 + ℎ) −〖 sin〗⁡〖 ( 𝑥 + 1 )〗)/h〗 Using sin A – sin B = 2 cos ((𝐴 + 𝐵)/2) sin ((𝐴 − 𝐵)/2) = lim┬(h→0)⁡〖(2 cos⁡(( ( 𝑥 + 1 + ℎ ) + ( 𝑥 + 1))/2).sin⁡((( 𝑥 + 1 + ℎ ) − ( 𝑥 + 1 ))/2))/h〗 = lim┬(h→0)⁡〖(2 cos⁡〖(( 2(𝑥 + 1 )+ ℎ))/2〗 . 〖sin 〗⁡〖ℎ/2〗)/h〗 = lim┬(h→0)⁡〖(cos⁡〖(( 2(𝑥 + 1 )+ ℎ))/2〗 . 〖sin 〗⁡〖ℎ/2〗)/(ℎ/2)〗 = lim┬(h→0)⁡〖〖cos 〗⁡〖(( 2(𝑥 + 1 )+ ℎ))/2〗 . sin⁡〖 ℎ/2〗/(( ℎ)/2)〗 = lim┬(h→0)⁡〖cos 〗⁡〖(( 2(𝑥 + 1 )+ ℎ))/2〗 .(𝐥𝐢𝐦)┬(𝐡→𝟎) 𝒔𝒊𝒏⁡〖 𝒉/𝟐〗/(( 𝒉)/𝟐) = lim┬(h→0)⁡〖cos 〗⁡〖(( 2(𝑥 + 1 ) + ℎ))/2〗 .𝟏 = lim┬(h→0)⁡〖cos 〗⁡〖(( 2(𝑥 + 1 )+ ℎ))/2〗 Putting h = 0 =〖 cos 〗⁡〖(( 2(𝑥 + 1 ) + 0))/2〗 = cos ( x + 1) Hence f’ (x) = cos (x + 1) Using (𝑙𝑖𝑚)┬(𝑥→0)⁡〖 𝑠𝑖𝑛⁡𝑥/𝑥〗=1 Replacing x by ℎ/2 ⇒ (𝑙𝑖𝑚)┬(ℎ→0) 𝑠𝑖𝑛⁡〖 ℎ/2〗/(( ℎ)/2) = 1