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प्रश्न
Differentiate \[\frac{x^2 + 2}{\sqrt{\cos x}}\] ?
उत्तर
\[\text{Let }y = \frac{x^2 + 2}{\sqrt{\cos x}}\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{\sqrt{\cos x}\frac{d}{dx}\left( x^2 + 2 \right) - \left( x^2 + 2 \right)\frac{d}{dx}\left( \sqrt{\cos x} \right)}{\left( \sqrt{\cos x} \right)^2} \left[ \text{Using quotient rule and chain rule} \right]\]
\[ = \frac{2x\sqrt{\cos x} - \left( x^2 + 2 \right)\left( - \frac{1}{2}\frac{\sin x}{\sqrt{\cos x}} \right)}{\cos x}\]
\[ = \frac{2x\sqrt{\cos x} + \frac{\left( x^2 + 2 \right)\sin x}{2\sqrt{\cos x}}}{\cos x}\]
\[ = \frac{4x \cos x + \left( x^2 + 2 \right)\sin x}{2 \left( \cos x \right)^\frac{3}{2}}\]
\[ = \frac{2x}{\sqrt{\cos x}} + \frac{1}{2}\frac{\left( x^2 + 2 \right)\sin x}{\left( \cos x \right)^\frac{3}{2}}\]
\[ = \frac{1}{\sqrt{\cos x}}\left\{ 2x + \frac{1}{2}\frac{\left( x^2 + 2 \right)\sin x}{\cos x} \right\}\]
\[ = \frac{1}{\sqrt{\cos x}}\left\{ 2x + \frac{\left( x^2 + 2 \right)\tan x}{2} \right\}\]
\[So, \frac{d}{dx}\left( \frac{x^2 + 2}{\sqrt{\cos x}} \right) = \frac{1}{\sqrt{\cos x}}\left\{ 2x + \frac{\left( x^2 + 2 \right)\tan x}{2} \right\}\]
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