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If $\sqrt{x}+\sqrt{y}=2$, then what is $\frac{d y}{d x}$ at $y=1$ and $\mathrm{x}=1$ equal to?
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Verified Answer
The correct answer is:
$-1$
Let $\sqrt{x}+\sqrt{y}=2$
Differentiate w.r.t.x, we get
$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$ ...(1)
Put $\mathrm{y}=1, \mathrm{x}=1$ in equation (1)
$\frac{1}{2}+\frac{1}{2} \frac{d y}{d x}=0$
$\Rightarrow \quad \frac{d y}{d x}=-1$
Differentiate w.r.t.x, we get
$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$ ...(1)
Put $\mathrm{y}=1, \mathrm{x}=1$ in equation (1)
$\frac{1}{2}+\frac{1}{2} \frac{d y}{d x}=0$
$\Rightarrow \quad \frac{d y}{d x}=-1$
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