Misc 1 - Find derivative by first principle: -x, sin (x+1) - Derivatives by 1st principle - At a general point

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  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise
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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﷯﷮ −𝑥 − ℎ + 𝑥﷮h﷯﷯ = lim﷮h→0﷯﷮ − ℎ﷮h﷯﷯ = lim﷮h→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) = lim﷮h→0﷯ f﷮ x + h﷯ − f(x)﷯﷮h﷯ Here, f (x) = −𝑥﷯﷮−1﷯ So, f (x + h) = −(𝑥+ℎ)﷯﷮−1﷯ Putting values f’ (x) = lim﷮h→0﷯﷮ − 𝑥 + ℎ﷯﷯ ﷮−1﷯ − (− 𝑥)﷮−1﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ 1﷮−(𝑥 + ℎ)﷯﷯ − 1﷮−𝑥﷯﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ − 1﷮𝑥 + ℎ﷯ + 1﷮𝑥﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ − 𝑥 + 𝑥 + ℎ﷮ℎ. 𝑥 (𝑥 + ℎ)﷯﷯ = lim﷮h→0﷯﷮ ℎ﷮ℎ𝑥 (𝑥 + ℎ)﷯﷯ = lim﷮h→0﷯﷮ 1﷮𝑥 (𝑥 + ℎ)﷯﷯ Putting h = 0 = 1﷮𝑥 (𝑥 + 0)﷯ = 𝟏﷮𝒙𝟐﷯ 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) Misc 1 Find the derivative of the following functions from first principle: (iv) cos x− π﷮8﷯﷯ Let f (x) = cos x− π﷮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− π﷮8﷯﷯ So, f (x + h) = cos (x+h)− π﷮8﷯﷯ Putting values f’(x) = lim﷮h→0﷯﷮ cos x + h﷯ − π﷮8﷯﷯ − cos x − π﷮8﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ cos 𝑥 − π﷮8﷯ +ℎ﷯ − cos x − π﷮8﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ −2 sin﷮ 𝑥 − 𝜋﷮8﷯ + ℎ﷯ + 𝑥 − 𝜋﷮8﷯ ﷯﷮2﷯﷯ . sin﷮ 𝑥 − 𝜋﷮8﷯ + ℎ﷯ − 𝑥 − 𝜋﷮8﷯ ﷯﷮2﷯﷯﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ −2 sin﷮ 2 𝑥 − 𝜋﷮8﷯ + ℎ﷯ ﷮2﷯﷯ . sin﷮ ℎ﷮2﷯﷯﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ − sin﷮ 2 𝑥 − 𝜋﷮8﷯ ﷯ + ℎ ﷮2﷯﷯ . sin﷮ ℎ﷮2﷯﷯﷯﷯﷯﷮ ℎ﷮2﷯﷯﷯ = lim﷮h→0﷯﷮ − sin ﷮ 2 𝑥 − 𝜋﷮8﷯ ﷯ + ℎ﷯﷮2﷯﷯ . sin﷮ ℎ﷮2﷯﷯﷮ ℎ﷮2﷯﷯﷯ = lim﷮h→0﷯﷮ − sin ﷮ 2 𝑥 − 𝜋﷮8﷯ ﷯+ ℎ﷯﷮2﷯﷯ . 𝐥𝐢𝐦﷮𝐡→𝟎﷯ 𝒔𝒊𝒏﷮ 𝒉﷮𝟐﷯﷯﷮ 𝒉﷮𝟐﷯﷯﷯ = lim﷮h→0﷯﷮ − sin ﷮ 2 𝑥 − 𝜋﷮8﷯ ﷯+ ℎ﷯﷮2﷯﷯ .𝟏﷯ = lim﷮h→0﷯﷮ − sin ﷮ 2 𝑥 − 𝜋﷮8﷯ ﷯+ ℎ﷮2﷯﷯﷯ ﷯ Putting h = 0 = − sin ﷮ 2 𝑥 − 𝜋﷮8﷯ ﷯+ 0﷮2﷯﷯﷯ = − sin ﷮ 2 𝑥 − 𝜋﷮8﷯ ﷯﷮2﷯﷯﷯ = − 𝐬𝐢𝐧 ﷮ 𝒙 − 𝝅﷮𝟖﷯ ﷯﷯

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