[tex] \sqrt \: sinx^{2} [/tex]
Find dy/dx of sqrt of sinx^2
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[tex] \sqrt \: sinx^{2} [/tex]
Find dy/dx of sqrt of sinx^2
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[tex]\large\underline{\sf{Given- }}[/tex]
[tex]\rm :\longmapsto\:y = \sqrt{sin {x}^{2} } [/tex]
[tex]\large\underline{\sf{To\:Find - }}[/tex]
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\rm :\longmapsto\:y = \sqrt{sin {x}^{2} } [/tex]
On differentiating both sides w. r. t. x we get
[tex]\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} \sqrt{sin {x}^{2} } [/tex]
We know,
[tex]\boxed{ \rm{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}[/tex]
So, using this, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{sin {x}^{2} } }\dfrac{d}{dx}sin {x}^{2} [/tex]
We know,
[tex]\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}[/tex]
So, using this, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{sin {x}^{2} } } \: \times cos {x}^{2} \dfrac{d}{dx} {x}^{2} [/tex]
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cos {x}^{2} }{2 \sqrt{sin {x}^{2} } } \: \dfrac{d}{dx} {x}^{2} [/tex]
We know,
[tex]\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}[/tex]
So, using this, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{cos {x}^{2} }{2 \sqrt{sin {x}^{2} } } \: \times 2x[/tex]
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{x \: cos {x}^{2} }{\sqrt{sin {x}^{2} } } \: [/tex]
Additional Information :-
[tex]\boxed{ \rm{ \dfrac{d}{dx}cosx = - sinx}}[/tex]
[tex]\boxed{ \rm{ \dfrac{d}{dx}tanx = {sec}^{2} x}}[/tex]
[tex]\boxed{ \rm{ \dfrac{d}{dx}cotx = - {cosec}^{2} x}}[/tex]
[tex]\boxed{ \rm{ \dfrac{d}{dx}secx = secx \: tanx}}[/tex]
[tex]\boxed{ \rm{ \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}[/tex]
[tex]\boxed{ \rm{ \dfrac{d}{dx}logx = \frac{1}{x}}}[/tex]
[tex]\boxed{ \rm{ \dfrac{d}{dx}k = 0}}[/tex]
[tex]\boxed{ \rm{ \dfrac{d}{dx}x = 1}}[/tex]