1. Class 11
2. Important Question for exams Class 11
3. Chapter 13 Class 11 Limits and Derivatives

Transcript

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

Chapter 13 Class 11 Limits and Derivatives

Class 11
Important Question for exams Class 11