Question

differentiate : 2√cot (x²)​

Answer 1

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

$\red{\rm :\longmapsto\:y = 2 \sqrt{cot {x}^{2} }}$

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

${\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}2 \sqrt{cot {x}^{2} }}$

We know,

$\red{\rm :\longmapsto\:\dfrac{d}{dx}k \: f(x) = k\dfrac{d}{dx} \: f(x)}$

So, using this we get

${\rm :\longmapsto\:\dfrac{d}{dx}y =2 \: \dfrac{d}{dx}\sqrt{cot {x}^{2} }}$

Now, we know,

$\red{\rm :\longmapsto\:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2 \times \dfrac{1}{2 \sqrt{cot {x}^{2} } } \dfrac{d}{dx}cot {x}^{2}$

We know,

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cotx = - {cosec}^{2}x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{cot {x}^{2} } } ( - {cosec}^{2} {x}^{2} )\dfrac{d}{dx} {x}^{2}$

We know,

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{cot {x}^{2} } } ( - {cosec}^{2} {x}^{2} )(2x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: \dfrac{1}{\sqrt{cot {x}^{2} } } ({cosec}^{2} {x}^{2} )(2x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ -2 x \: {cosec}^{2} {x}^{2} }{\sqrt{cot {x}^{2} } }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - 2x \:\sqrt{sin {x}^{2} } }{ {sin}^{2} {x}^{2} \sqrt{cos {x}^{2} } }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ -2 x \:\sqrt{sin {x}^{2} } }{ sin {x}^{2} \: . \: {sin} {x}^{2} \sqrt{cos {x}^{2} } }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ -2 x \: }{ sin {x}^{2} \: . \: \sqrt{sin {x}^{2} } \sqrt{cos {x}^{2} } }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - 2\sqrt{2} x \: }{ sin {x}^{2} \: \sqrt{2} \: \sqrt{sin {x}^{2} } \sqrt{cos {x}^{2} } }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ -2 \sqrt{2} x \: }{ sin {x}^{2} \: \sqrt{2\:sin {x}^{2} cos {x}^{2} }}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - 2\sqrt{2} x \: }{ sin {x}^{2} \: \sqrt{\:sin2 {x}^{2} }}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - 2\sqrt{2} x \: cosec {x}^{2} \: }{ \: \sqrt{\:sin2 {x}^{2} }}$

Additional Information :-

$\boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cosx = - \: sinx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}secx = \: secx \: tanx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cotx = - \: {cosec}^{2}x}}$

Answer 2

Answer:

-2xcsc²x² √tan(x²)

Step-by-step explanation:

As per the information provided in the given question, We have :

• y = 2√cot (x²)

We are given with the value of y. We are asked to find d/dx with respect to y i.e 2√cot (x²).

In order to find d/dx, We will use chain rule.

$\longmapsto \rm \dfrac{d}{dx} (2 \sqrt{ \cot( {x}^{2} ) } )$

This can be written as,

$\longmapsto \rm 2\dfrac{d}{dx} (\sqrt{ \cot( {x}^{2} ) } )$

Applying the chain rule,

$\longmapsto \rm \dfrac{1}{ 2\sqrt{ \cot( {x}^{2} ) } } \dfrac{d}{dx} \bigg ( \cot( {x}^{2} ) \bigg )$

d/dx of cot(x) is - csc²x Thus, d/dx of cot(x²) is - csc²x²,

$\longmapsto \rm 2 \: \dfrac{1}{ 2\sqrt{ \cot( {x}^{2} ) } } ( - { \csc}^{2} {x}^{2} ) \times 2x$

By cancelling 2,

$\longmapsto \rm \: \dfrac{1}{ \sqrt{ \cot( {x}^{2} ) } } ( - { \csc}^{2} {x}^{2} ) \times 2x$

1/√cot(x) is √tan x. Thus, 1/√cot(x²) is √tan x²,

$\longmapsto \rm \: { \sqrt{ \tan( {x}^{2} ) } } ( - { \csc}^{2} {x}^{2} ) \times 2x$

$\longmapsto \rm \: - 2x { \csc}^{2}{x}^{2} { \sqrt{ \tan( {x}^{2} ) } }$

∴ -2xcsc²x²√tan(x²) is the derivative of 2√cot (x²).

Learn more!

$\begin{gathered}\boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\text{Differentiation \:formulae:}}\\\\\bigstar\:\:\sf{ \dfrac{d}{dx} \times { {x}^{n}} = n \times {x}^{n - 1} } \\\\\bigstar\:\:\sf{ \dfrac{d}{dx} ( {x}) = 1 }\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( { Cons}) = 0\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( { log(x) }) = \dfrac{1}{x} \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sqrt{x}} )= \dfrac{1}{2 \sqrt{x} }\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sin(x)} )= \cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\cos(x)} )= - \sin(x) \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sin(x)} )= \cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\tan(x)} )= { \sec}^{2} (x) \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\cot(x)} )=- { \cosec}^{2} (x) \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sec(x)} )=\sec(x) \times \tan(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\cosec(x)})= - \cosec(x) \times \cot(x)\end{array}}\end{gathered}$