Differentiate the following functions with respect to x: x^2+2 - Vidyadip

Question

Differentiate the following functions with respect to x:
$\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$

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Answer

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

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