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Let $y= \sqrt{\cos (2x+3)}$ then dy/dx equals

Option 1)
$\frac{- \sin (2x+3)}{\sqrt{\cos (2x+3)}}$

Option 2)
$\frac{ \sin (2x+3)}{\sqrt{\cos (2x+3)}}$

Option 3)
$\sqrt{ -\sin (2x+3)}}$

Option 4)
$\sqrt {-2 \sin(2x+3) }$

Answers (1)

best_answer

As we have learned

Chain Rule for differentiation (indirect) -

$Let\:\:y=\sqrt{sin(ax+b)}$

$Let\:\:y=sin(ax+b)=u$

$and\:\(ax+b)=v$

$then\;\:y=\sqrt{u},\;\:u=sinv$

$\therefore \:\:\frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}$

$=\frac{1}{2\sqrt{u}}\times cosv\times a$

$=\frac{a}{2}.\frac{cosv}{\sqrt{u}}=\frac{a\:cos(ax+b)}{2\sqrt{sin(ax+b)}}$

- wherein

$\because \:\:\frac{d}{dx}\sqrt{x}<\frac{1}{2\sqrt{x}}$

Let $2x+3=u$ and $\cos (2x+3) =v$

$\therefore y=\sqrt v$ where $v= \cos u$

$\frac{dy}{dx}=\frac{dy}{dv}\cdot \frac{dv}{dx}=\frac{du}{dx}= \frac{d\sqrt{v}}{dv}\cdot \frac{d(\cos u)}{du}\cdot \frac{d(2x+3)}{dx}$

$\Rightarrow \frac{dy}{dx}=(1/2 \sqrt v)\cdot (-\sin u )\cdot 2= \frac{-\sin (2x+3)}{\sqrt{\cos (2x+3)}}$

Option 1)

$\frac{- \sin (2x+3)}{\sqrt{\cos (2x+3)}}$

Option 2)

$\frac{ \sin (2x+3)}{\sqrt{\cos (2x+3)}}$

Option 3)

$\sqrt{ -\sin (2x+3)}}$

Option 4)

$\sqrt {-2 \sin(2x+3) }$

Posted by

Himanshu

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