Question

$\sqrt \: sinx^{2}$
Find dy/dx of sqrt of sinx^2
​

Answer 1

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

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{dy}{dx}$

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

Given that

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

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

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

We know,

$\boxed{ \rm{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}$

So, using this, we get

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

We know,

$\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}$

So, using this, we get

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

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

We know,

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

So, using this, we get

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

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

Additional Information :-

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

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

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

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

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

$\boxed{ \rm{ \dfrac{d}{dx}logx = \frac{1}{x}}}$

$\boxed{ \rm{ \dfrac{d}{dx}k = 0}}$

$\boxed{ \rm{ \dfrac{d}{dx}x = 1}}$