Example 20 - Chapter 13 Class 11 Limits and Derivatives

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Example 20 Find the derivative of f(x) from the first principle, where f(x) is (i) sin x + cos x Given f (x) = sin x + cos x 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 + cos x So, f (x + h) = sin (x + h) + cos (x + h) Putting values f’(x) = lim﷮h→0﷯﷮ sin﷮(𝑥 + ℎ)﷯ + cos﷮ 𝑥 + ℎ﷯﷯ − ( sin﷮𝑥﷯ + cos﷮𝑥)﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ + cos﷮𝑥 cos﷮ℎ − sin﷮𝑥 sin﷮ℎ − sin﷮𝑥 − cos﷮𝑥﷯﷯﷯﷯﷯ ﷯﷯﷯﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ cos﷮𝑥 sin﷮ℎ − sin﷮𝑥 sin﷮ℎ + sin﷮𝑥 cos﷮ℎ − sin﷮𝑥 + cos﷮𝑥 cos﷮ℎ − cos﷮𝑥﷯﷯﷯﷯﷯ ﷯﷯﷯﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ sin﷮ℎ (cos﷮𝑥 − sin﷮𝑥) + sin﷮𝑥 ( cos﷮ℎ − 1) + cos⁡𝑥 ( cos﷮ℎ − 1)﷯﷯﷯﷯﷯﷯﷮h﷯﷯ = lim﷮h→0﷯﷮ sin﷮ℎ (cos﷮𝑥 − sin﷮𝑥)﷯﷯﷯﷮h﷯+ sin﷮𝑥 ( cos﷮ℎ − 1)﷯﷯﷮h﷯+ cos﷮𝑥 ( cos﷮ℎ − 1)﷯﷯﷮ℎ﷯﷯ ﷯ = lim﷮h→0﷯﷮ sin﷮ℎ (cos﷮𝑥 − sin﷮𝑥)﷯﷯﷯﷮h﷯+ lim﷮h→0﷯ sin﷮𝑥 ( cos﷮ℎ − 1)﷯﷯﷮h﷯+ lim﷮h→0﷯ cos﷮𝑥 ( cos﷮ℎ − 1)﷯﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮(cos x – sin x) sin﷮ℎ﷯﷮ℎ﷯+ lim﷮h→0﷯(– sin x) (1 − cos﷮ℎ)﷯﷮ℎ﷯﷯+ lim﷮h→0﷯(– cos x) (1 − cos﷮ℎ)﷯﷮ℎ﷯ = (cos x – sin x) 𝐥𝐢𝐦﷮𝐡→𝟎﷯﷮ 𝐬𝐢𝐧﷮𝒉﷯﷮𝒉﷯− sin﷮𝑥 𝐥𝐢𝐦﷮𝐡→𝟎﷯﷯ (𝟏 − 𝒄𝒐𝒔﷮𝒉)﷯﷮𝒉﷯− cos﷮𝑥﷯ 𝐥𝐢𝐦﷮𝐡→𝟎﷯ (𝟏 − 𝒄𝒐𝒔﷮𝒉)﷯﷮𝒉﷯﷯ = (cos x – sin x) × 1 – sin﷮𝑥 ﷯ × (0) – cos﷮𝑥﷯ × (0) = cos x – sin x – 0 – 0 = cos x – sin x Hence, f’(x) = cos x – sin x Example, 20 Find the derivative of f(x) from the first principle, where f(x) is (ii) x sin x Given f (x) = x sin x 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) = x sin x So, f (x + h) = (x + h) sin (x + h) Putting values f’(x) = lim﷮h→0﷯ 𝑥 + ℎ﷯ sin﷮ 𝑥 + ℎ﷯ − 𝑥 sin﷮𝑥 ﷯﷯﷮ℎ﷯ Using sin (A + B) = sin A cos B + cos A sin B = lim﷮h→0﷯﷮ 𝑥+ℎ﷯( sin﷮𝑥 cos﷮ℎ + cos 𝑥﷮ sin﷮ℎ ﷯﷯)﷯ − 𝑥 sin﷮𝑥﷯﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ 𝑥( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ) + ℎ ( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ) − 𝑥 sin﷮𝑥﷯﷯﷯﷯﷯﷯﷯﷯﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ 𝑥 sin﷮𝑥 cos﷮ℎ + 𝑥 cos﷮𝑥 sin﷮ℎ + ℎ (sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ) − 𝑥 sin﷮𝑥﷯﷯﷯﷯﷯﷯﷯﷯﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ 𝑥𝑠𝑖𝑛 𝑥 cos﷮ℎ − 𝑥 sin﷮𝑥 + 𝑥 cos﷮𝑥 sin﷮ℎ + ℎ( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ﷯﷯﷯)﷯﷯﷯﷯﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ 𝑥𝑠𝑖𝑛 𝑥 (cos﷮ℎ − 1)+ 𝑥 cos﷮𝑥 sin﷮ℎ + ℎ( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ﷯﷯﷯)﷯﷯ ﷯﷯﷮ℎ﷯﷯ = lim﷮h→0﷯﷮ 𝑥 sin﷮𝑥 ( cos﷮ℎ −1)﷯﷯﷮ℎ﷯+ 𝑥 cos﷮𝑥 sin﷮ℎ﷯﷯﷮ℎ﷯+ ℎ ( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ)﷯﷯﷯﷯﷮ℎ﷯﷯﷯ = lim﷮h→0﷯﷮ x sin﷮𝑥 ( cos﷮ℎ −1)﷯﷯﷮ℎ﷯+ lim﷮h→0﷯ 𝑥 cos﷮𝑥 sin﷮ℎ﷯﷯ ﷮h﷯+ lim﷮h→0﷯( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ )﷯﷯﷯﷯﷯ = lim﷮h→0﷯﷮𝑥 sin﷮𝑥 ( cos﷮ℎ −1)﷯﷮ℎ﷯﷯+ lim﷮h→0﷯ 𝑥 cos﷮𝑥﷯ sin﷮ℎ﷯﷮h﷯+ lim﷮h→0﷯( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ )﷯﷯﷯﷯﷯ = – x sin x 𝐥𝐢𝐦﷮𝐡→𝟎﷯ (𝟏− 𝒄𝒐𝒔﷮𝒉)﷯﷮𝒉﷯+𝑥 cos﷮𝑥 𝐥𝐢𝐦﷮𝐡→𝟎﷯ 𝒔𝒊𝒏﷮𝒉﷯﷮𝐡﷯+﷯ lim﷮h→0﷯( sin﷮𝑥 cos﷮ℎ + cos﷮𝑥 sin﷮ℎ )﷯﷯﷯﷯ = – x sin x (0) + x cos x (1) + ( sin x cos 0 + cos x sin 0) = 0 + x cos x + sin × 1 + cos x × 0 = 0 + x cos x + sin x + 0 = x cos x + sin x Hence f’ (x) = x cos x + sin x