Question

Differentiate the function with respect to x
$\rm \: \implies2 \sqrt{ \cot( {x}^{2} ) }$
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Answer 1

Given :

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

To find :

dy/dx

Solution :

$⟹ \frac{d(2 \sqrt{ \cot( {x}^{2} ) }) }{dx}$

Now differentiate using chain rule :-

d[f(g(x))]/dx = f'(g(x)) g'(x)

$⟹ 2 \times \frac{d( \sqrt{ \cot( {x}^{2} ) }) }{dx}$

$⟹ - 2 \times \frac{1 }{2} \times \frac{1}{ \sqrt{cot \: {x}^{2} } } \times {cosec}^{2} x \times 2x$

$⟹ - \frac{1}{ \sqrt{cot \: {x}^{2} } } \times 2x \: {cosec}^{2} {x}^{2}$

$⟹ - \frac{2x \: {cosec}^{2} {x}^{2} }{ \sqrt{cot \: {x}^{2} } }$

we can write :- cot x² = cos x²/ sin x²

and cosec²x² =( 1/sin²x²)

$⟹ - \frac{2x }{ {sin}^{2} {x}^{2} \sqrt{cos {x}^{2} / sin {x}^{2} } }$

$⟹ - \frac{2x }{ {sin}^{2} {x}^{2} \sqrt{cos {x}^{2} sin {x}^{2} } }$

multiply both numerator and denominator by √2

therefore we get :-

$- \frac{2√2x}{ sin {x}^{2}\sqrt{sin{2x}^{2} } }$

Answer :-

$- \frac{2√2x}{ sin {x}^{2}\sqrt{sin{2x}^{2} } }$

( also check attachment)

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Learn more :-

• d(sinx)/dx = cosx
• d(cos x)/dx = -sin x
• d(cosec x)/dx = -cot x cosec x
• d(tan x)/dx = sec²x
• d(sec x)/dx = secx tanx
• d(cot x)/dx = - cosec² x

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Answer 2

Given

$\bullet\ \; \bf 2\sqrt{cot(x^2)}$

To Find

Derivative of the given value

Formula Applied

$\bullet\ \; \rm \dfrac{d}{dx}(cotx)=-csc^2x\\\\\bullet\ \; \rm sin2x=2.sinx.cosx\\\\\bullet\ \; \rm \dfrac{d}{dx}(\sqrt{x})=\dfrac{1}{2\sqrt{x}}$

Solution

Let ,

$\bf y=2\sqrt{cot(x^2)}$

Differentiating with respect to x on both sides ,

$\to \rm \dfrac{dy}{dx}=\dfrac{d}{dx}(2\sqrt{cot(x^2)})\\\\\to \rm \dfrac{dy}{dx}=\dfrac{2}{2\sqrt{cot(x^2)}}.\dfrac{d}{dx}(cot(x^2))\\\\\to \rm \dfrac{dy}{dx}=\dfrac{1}{\sqrt{cot(x^2)}}(-csc(x^2))(2x)\\\\\to \rm \dfrac{dy}{dx}=\dfrac{-2x.csc^2(x^2)}{\sqrt{cot(x^2)}}\\\\\to \rm \dfrac{dy}{dx}=\dfrac{-2x}{sin^2(x^2)\sqrt{\frac{cos(x^2)}{sin(x^2)}}}\\\\\to \rm \dfrac{dy}{dx}=\dfrac{-2x}{sin(x^2)\sqrt{sin(x^2)cos(x^2)}}\\\\\to \bf \pink{\dfrac{dy}{dx}=\dfrac{-2\sqrt{2}x}{sin(x^2)\sqrt{sin(2x^2)}}\ \; \bigstar}$