Question

Differentiate with respect to x
d 2√cotx^2
dx​

Answer 1

EXPLANATION.

$\sf \implies 2\sqrt{cot(x^{2} )}$

As we know that,

Let, $\sf \implies y = 2\sqrt{cot(x^{2} )}$

Differentiate w.r.t x, we get.

$\sf \implies \dfrac{dy}{dx} = \dfrac{d(2\sqrt{cot(x^{2} )} }{dx}$

$\sf \implies \dfrac{dy}{dx} = 2 \ \times \dfrac{1}{2\sqrt{cot(x^{2} )} } \ \times \dfrac{d(cot(x^{2} ))}{dx}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{1}{\sqrt{cot(x^{2} )} } \ \times -2x (cosec^{2} x^{2} )$

$\sf \implies \dfrac{dy}{dx} = \dfrac{-2x(cosec^{2}x^{2} ) }{\sqrt{cot(x^{2} )} }$

$\sf \implies \dfrac{dy}{dx} =\dfrac{-2x}{\sqrt{\dfrac{cos(x^{2} )}{sin(x^{2} )} } } \ \times \dfrac{1}{sin^{2} (x^{2} )}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{-2x}{sin(x)^{2} \times \sqrt{cos(x^{2} ).sin(x^{2} ).} }$

$\sf \implies \dfrac{dy}{dx} = \dfrac{-2x}{sin(x)^{2} \times \sqrt{cos(x^{2} ).sin(x^{2} ).} } \ \times \sqrt{\dfrac{2}{2} }$

$\sf \implies \dfrac{dy}{dx} = \dfrac{-2\sqrt{2} x}{sin(x)^{2} \times \sqrt{2 \times cos(x^{2} ).sin(x^{2} ).} }$

$\sf \implies \dfrac{dy}{dx} = \dfrac{-2\sqrt{2} x}{sin(x)^{2} \times \sqrt{sin(2x^{2} ).} }$

                                                                                                                         

MORE INFORMATION.

(1) = d(sin x)/dx = cos x.

(2) = d(cos x)/dx = - sin x.

(3) = d(tan x)/dx = sec²x.

(4) = d(cot x)/dx = -cosec²x.

(5) = d(sec x)/dx = sec x tan x.

(6) = d(cosec x)/dx = -cosec x cot x.

Answer 2

$let \: \: \: \: y = 2 \sqrt{cot( {x}^{2}) } \\ \\ \frac{dy}{dx} = 2 \frac{1}{2 \sqrt{cot( {x}^{2}) } } \frac{d}{dx} cot ({x}^{2} ) \\ \\ \frac{dy}{dx} = \frac{1}{ \sqrt{cot {x}^{2} } } ( - cosec ^{2} {x}^{2} ) \frac{d}{dx} {x}^{2} \\ \\ \frac{dy}{dx} = \frac{ - cosec ^{2} {x}^{2} }{ \sqrt{cot {x}^{2} } } 2x \\ \\ \frac{dy}{dx} = \frac{ - 2x {cosec}^{2} {x}^{2} }{ \sqrt{cot {x}^{2} } }$