Last updated at March 9, 2017 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) = limh→0 f x + h − f(x)h Here, f (x) = – x So, f (x + h) = – (x + h) Putting values f’ (x) = limh→0 − x + h − (− x)h = limh→0 −𝑥 − ℎ + 𝑥h = limh→0 − ℎh = limh→0(−1) = – 1 Hence, f’(x) = – 1 Misc 1 Find the derivative of the following functions from first principle: (ii) −𝑥−1 Let f (x) = −𝑥−1 We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = limh→0 f x + h − f(x)h Here, f (x) = −𝑥−1 So, f (x + h) = −(𝑥+ℎ)−1 Putting values f’ (x) = limh→0 − 𝑥 + ℎ −1 − (− 𝑥)−1ℎ = limh→0 1−(𝑥 + ℎ) − 1−𝑥ℎ = limh→0 − 1𝑥 + ℎ + 1𝑥ℎ = limh→0 − 𝑥 + 𝑥 + ℎℎ. 𝑥 (𝑥 + ℎ) = limh→0 ℎℎ𝑥 (𝑥 + ℎ) = limh→0 1𝑥 (𝑥 + ℎ) Putting h = 0 = 1𝑥 (𝑥 + 0) = 𝟏𝒙𝟐 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) = limh→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) = limh→0 sin 𝑥 + ℎ + 1 − sin (𝑥 + 1)h = limh→0 𝑠𝑖𝑛 𝑥 + 1 + ℎ − sin ( 𝑥 + 1 )h Using sin A – sin B = 2 cos 𝐴 + 𝐵2 sin 𝐴 − 𝐵2 = limh→0 2 cos 𝑥 + 1 + ℎ + 𝑥 + 12. sin 𝑥 + 1 + ℎ − ( 𝑥 + 1 )2h = limh→0 2 cos ( 2 𝑥 + 1 + ℎ)2 . sin ℎ2h = limh→0 cos ( 2 𝑥 + 1 + ℎ)2 . sin ℎ2 ℎ2 = limh→0 cos ( 2 𝑥 + 1 + ℎ)2 . sin ℎ2 ℎ2 = limh→0 cos ( 2 𝑥 + 1 + ℎ)2 . 𝐥𝐢𝐦𝐡→𝟎 𝒔𝒊𝒏 𝒉𝟐 𝒉𝟐 = limh→0 cos ( 2 𝑥 + 1 + ℎ)2 .𝟏 = limh→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) Misc 1 Find the derivative of the following functions from first principle: (iv) cos x− π8 Let f (x) = cos x− π8 We need to find Derivative of f(x) We know that f’(x) = limh→0 f x + h − f(x)h Here, f (x) = cos x− π8 So, f (x + h) = cos (x+h)− π8 Putting values f’(x) = limh→0 cos x + h − π8 − cos x − π8h = limh→0 cos 𝑥 − π8 +ℎ − cos x − π8h = limh→0 −2 sin 𝑥 − 𝜋8 + ℎ + 𝑥 − 𝜋8 2 . sin 𝑥 − 𝜋8 + ℎ − 𝑥 − 𝜋8 2h = limh→0 −2 sin 2 𝑥 − 𝜋8 + ℎ 2 . sin ℎ2h = limh→0 − sin 2 𝑥 − 𝜋8 + ℎ 2 . sin ℎ2 ℎ2 = limh→0 − sin 2 𝑥 − 𝜋8 + ℎ2 . sin ℎ2 ℎ2 = limh→0 − sin 2 𝑥 − 𝜋8 + ℎ2 . 𝐥𝐢𝐦𝐡→𝟎 𝒔𝒊𝒏 𝒉𝟐 𝒉𝟐 = limh→0 − sin 2 𝑥 − 𝜋8 + ℎ2 .𝟏 = limh→0 − sin 2 𝑥 − 𝜋8 + ℎ2 Putting h = 0 = − sin 2 𝑥 − 𝜋8 + 02 = − sin 2 𝑥 − 𝜋8 2 = − 𝐬𝐢𝐧 𝒙 − 𝝅𝟖

Misc 1
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Misc 2

Misc 3

Misc 4

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Misc 6 Important

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Misc 9 Important

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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 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.