Question

differentiate √tan x​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: \sqrt{tanx}$

Let assume that

$\rm \: f(x) = \sqrt{tanx}$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}f(x) =\dfrac{d}{dx} \sqrt{tanx}$

can be rewritten as

$\rm \: f'(x) = \dfrac{d}{dx} {\bigg(tanx \bigg) }^{\dfrac{1}{2} }$

We know,

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

So, using this result, we get

$\rm \: f'(x) = \dfrac{1}{2} {\bigg(tanx \bigg) }^{\dfrac{1}{2} - 1}\dfrac{d}{dx}tanx$

$\rm \: f'(x) = \dfrac{1}{2} {\bigg(tanx \bigg) }^{ - \dfrac{1}{2}} {sec}^{2}x$

$\rm \: f'(x) = \dfrac{1}{2{\bigg(tanx \bigg) }^{\dfrac{1}{2}}} {sec}^{2}x$

$\rm\implies \:\rm \: f'(x) =\dfrac{1}{2 \sqrt{tanx} } \times {sec}^{2}x$

Or

$\rm\implies \:\rm \: f'(x) =\dfrac{ {sec}^{2} x}{2 \sqrt{tanx} }$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

Use Chain Rule on $\displaystyle \rm\frac{d}{dx} \sqrt{\tan{x}}$ Let u=tanx.

Since $\rm\sqrt{u}={u}^{\frac{1}{2}}$ , using the Power Rule, $\displaystyle\rm\frac{d}{du} {u}^{\frac{1}{2}}=\frac{1}{2}{u}^{-\frac{1}{2}}$

$\\ \implies\rm\pmb{\frac{1}{2\sqrt{\tan{x}}} \bigg(\frac{d}{dx} \tan{x} \bigg)}$

$\\ \colorbox{olive}{ \boxed{ \red{\implies\rm { \frac{ \sec {}^{2} (x) }{2 \sqrt{ \tan(x) } } }}}}$