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Compute the derivative of f(x) = cosx. |
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Answer» \(\frac d{dx}cosx=\lim\limits_{h \to 0}\frac{cos(x+h)-cos x}h\) \((\because f^1(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}h)\) \(=\lim\limits_{h\to 0}\frac{cos xcos h-sin xsin h-cos x}h\) (\(\because\) cos (x + y) = cos x cos y - sin x sin y) \(=\lim\limits_{h\to 0}\frac{cosx(cos h-1)}h-\lim\limits_{h\to 0}sin x\frac{sin h}h\) \(=cos x\lim\limits_{h \to x}\cfrac{1-sin^2\frac h2-1}{\frac{h^2}4\times\frac 4h}-sin x\lim\limits_{h\to 0}\frac{sin h}h\) \(=-\frac24cos x\lim\limits_{h \to 0}h\left(\cfrac{sin\frac h2}{\frac h2}\right)^2\) - sin x \(\lim\limits_{h\to 0}\frac{sin h}h\) \(=-\frac12cos x\lim\limits_{h\to 0}h\left(\lim\limits_{h/2 \to 0}\cfrac{sin\frac h2}{\frac h2}\right)^2\) - sin x \(\lim\limits_{h\to 0}\frac{sin h}h\) \(=-\frac12 cos x\times0\times1^2-sin x\times1\) (\(\because\lim\limits_{h\to 0}\frac{sin h}h=1\)) = 0 - sin x = - sin x Hence, \(\frac d{dx} cos x=- sin x\) |
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