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If $y=\sqrt{sin x+y}$, then $\Large \frac{dy}{dx}$ is equal to

\[ \begin{array}{1 1}(A)\;\frac{\cos x}{2y-1} & (B)\;\frac{\cos x}{1-2y}\\(C)\;\frac{\sin x}{1-2y} & (D)\;\frac{\sin x}{2y-1}\end{array}\]

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  • If the variable $x$ and $y$ are connected by a relation of the form $f(x,y)=0$ and it is not possible to express $y$ as a function of $x$,then it is said to be implicit function.
  • Hence $\phi(y)$ w.r.t $x$ is $\large\frac{d\phi}{dy}.\frac{dy}{dx}$
Step 1:
$y=\sqrt{\sin x+y}$
$\large\frac{dy}{dx}=\large\frac{1}{2}$$(\sin x+y)^{\Large\frac{-1}{2}}(\cos x+\large\frac{dy}{dx})$
$\Rightarrow \large\frac{1}{2\sqrt{\sin x+y}}\times$$ \big(\cos x+\large\frac{dy}{dx}\big)$
$\large\frac{dy}{dx}$$\big(1-\large\frac{1}{2\sqrt{\sin x+y}}\big)=\large\frac{\cos x}{2\sqrt{\sin x+y}}$
$\Rightarrow \large\frac{dy}{dx}=\large\frac{\Large\frac{\cos x}{2\sqrt{\sin x+y}}}{1-\Large\frac{1}{2\sqrt{\sin x+y}}}$
Step 2:
On simplifying we get,
$\large\frac{dy}{dx}=\large\frac{\cos x}{2\sqrt{\sin x+y}-1}$
But $y=\sqrt{\sin x+y}$
$\qquad=\large\frac{\cos x}{2y-1}$
Hence $A$ is the correct option.
answered Jul 4, 2013 by sreemathi.v

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