Differentiate x w.r.t. x from first principles. - Vidyadip

Question

Differentiate $\sqrt{\tan\text{ x}}$ w.r.t. x from first principles.

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Answer

$\frac{\text{dy}}{\text{dx}}=\lim\limits_{\Delta\rightarrow0}\frac{\sqrt{\tan(\text{x +}\Delta\text{x})}-\sqrt{\tan\text{x}}}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x}\rightarrow0}\frac{1}{\Big(\sqrt{\tan\text{( x + }\Delta\text{x})}+\sqrt{\tan\text{x}}\Big)}\cdot\frac{[\tan\text{(x +}\Delta\text{x})-\tan\text{x}]}{\Delta\text{x}}$
$=\frac{1}{2\sqrt{\tan\text{x}}}\lim\limits_{\Delta\text{x}\rightarrow0}\frac{\tan\text{(x +}\Delta\text{x - x})[1+\tan\text{ (x +}\Delta\text{x}).\tan\text{x}]}{\Delta\text{x}}$
$=\frac{1}{2\sqrt{\tan\text{x}}}.1.(1+\tan^{2}\text{x})=\frac{\sec^{2}\text{x}}{2\sqrt{\tan\text{x}}}.$

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