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Encoding and Decoding Patterns

Differentiate...

Question

Differentiate using first principle x²cos x.

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Solution

Dear student
$Let f (x) = x^{2} cosx . Then, f (x + h) = {(x + h)}^{2} \cos (x + h) . \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{{(x + h)}^{2} \cos (x + h) - x^{2} cosx}{h} \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{(x^{2} + 2 hx + h^{2}) \cos (x + h) - x^{2} cosx}{h} \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} \frac{\{x^{2} \cos (x + h) - x^{2} \cos\} + 2 hxcos (x + h) + h^{2} \cos (x + h)}{h} \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} [x^{2} \{\frac{\cos (x + h) - cosx}{h}\} + 2 xcos (x + h) + hcos (x + h)] \Rightarrow \frac{d}{dx} (f (x)) = \lim_{h \to 0} - 2 x^{2} \frac{\sin (x + \frac{h}{2}) \sin \frac{h}{2}}{h} + \lim_{h \to 0} 2 xcos (x + h) + \lim_{h \to 0} hcos (x + h) \Rightarrow \frac{d}{dx} (f (x)) = - x^{2} \lim_{h \to 0} \frac{\sin (x + \frac{h}{2}) \sin \frac{h}{2}}{\frac{h}{2}} + \lim_{h \to 0} 2 xcos (x + h) + \lim_{h \to 0} hcos (x + h) \Rightarrow \frac{d}{dx} (f (x)) = - x^{2} sinx + 2 xcosx + 0 \times cosx \Rightarrow \frac{d}{dx} (f (x)) = - x^{2} sinx + 2 xcosx$
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