Question

find the derivate of √sin 2x​

Answer 1

Solution:

Given That:

$\rm \longrightarrow y = \sqrt{ \sin(2x) }$

We have to calculate derivative of y wrt x.

$\rm \longrightarrow {y}^{2} = \sin(2x)$

$\rm \longrightarrow {y}^{2} = 2\sin(x) \cos(x)$

Differentiating both sides wrt x, we get:

$\rm \longrightarrow 2y \dfrac{dy}{dx} = 2 \dfrac{d}{dx} \{\sin(x) \cos(x) \}$

$\rm \longrightarrow y \dfrac{dy}{dx} = \dfrac{d}{dx} \{\sin(x) \cos(x) \}$

$\rm \longrightarrow y \dfrac{dy}{dx} = \sin(x) \dfrac{d}{dx}\cos(x) + \cos(x) \dfrac{d}{dx} \sin(x)$

$\rm \longrightarrow y \dfrac{dy}{dx} = \sin(x) \cdot \sin(x) + \cos(x) \cdot - \sin(x)$

$\rm \longrightarrow y \dfrac{dy}{dx} = \sin^{2} (x) - \cos^{2} (x)$

$\rm \longrightarrow y \dfrac{dy}{dx} = \cos(2x)$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{ \cos(2x) }{y}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{ \cos(2x) }{ \sqrt{ \sin(2x) } }$

★ Which is our required answer.

Answer:

$\rm \hookrightarrow \dfrac{d}{dx} \sqrt{ \sin(2x) } = \dfrac{ \cos(2x) }{ \sqrt{ \sin(2x) } }$

Note:

$\rm\longrightarrow \sin(2x) = 2\sin(x)\cos(x)$

$\rm\longrightarrow \cos(2x) = \cos^{2}(x) - \sin^{2}(x)$

Additional Information:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$