Last updated at May 29, 2018 by Teachoo

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Misc 1 Find the derivative of the following functions from first principle: x Let f (x) = x We need to find derivative of f(x) i.e. f (x) We know that f (x) = lim h 0 f x + h f(x) h Here, f (x) = x So, f (x + h) = (x + h) Putting values f (x) = lim h 0 x + h ( x) h = lim h 0 + h = lim h 0 h = lim h 0 ( 1) = 1 Hence, f (x) = 1 Misc 1 Find the derivative of the following functions from first principle: (ii) 1 Let f (x) = 1 We need to find Derivative of f(x) i.e. f (x) We know that f (x) = lim h 0 f x + h f(x) h Here, f (x) = 1 So, f (x + h) = ( + ) 1 Putting values f (x) = lim h 0 + 1 ( ) 1 = lim h 0 1 ( + ) 1 = lim h 0 1 + + 1 = lim h 0 + + . ( + ) = lim h 0 ( + ) = lim h 0 1 ( + ) Putting h = 0 = 1 ( + 0) = 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 + + 1 sin ( + 1) h = lim h 0 + 1 + sin ( + 1 ) h Using sin A sin B = 2 cos + 2 sin 2 = lim h 0 2 cos + 1 + + + 1 2 . sin + 1 + ( + 1 ) 2 h = lim h 0 2 cos ( 2 + 1 + ) 2 . sin 2 h = lim h 0 cos ( 2 + 1 + ) 2 . sin 2 2 = lim h 0 cos ( 2 + 1 + ) 2 . sin 2 2 = lim h 0 cos ( 2 + 1 + ) 2 . = lim h 0 cos ( 2 + 1 + ) 2 . = lim h 0 cos ( 2 + 1 + ) 2 Putting h = 0 = cos ( 2 + 1 + 0) 2 = cos ( x + 1) Hence f (x) = cos (x + 1) Misc 1 Find the derivative of the following functions from first principle: (iv) cos x 8 Let f (x) = cos x 8 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) = cos x 8 So, f (x + h) = cos (x+h) 8 Putting values f (x) = lim h 0 cos x + h 8 cos x 8 h = lim h 0 cos 8 + cos x 8 h = lim h 0 2 sin 8 + + 8 2 . sin 8 + 8 2 h = lim h 0 2 sin 2 8 + 2 . sin 2 h = lim h 0 sin 2 8 + 2 . sin 2 2 = lim h 0 sin 2 8 + 2 . sin 2 2 = lim h 0 sin 2 8 + 2 . = lim h 0 sin 2 8 + 2 . = lim h 0 sin 2 8 + 2 Putting h = 0 = sin 2 8 + 0 2 = sin 2 8 2 =

Miscellaneous

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.