Question

If $\sqrt{x}+\sqrt{y}=4$, then find $\frac{dx}{dy}$ at $y=1$.

• A. $3$
• B. $4$
• C. $-3$
• D. $-4$

Answer 1

Answer: (C)

$-3$

$\sqrt{x}+\sqrt{y}=4$
Differentiating both sides of the given equation w.r.t. y, we get
$\frac{1}{2\sqrt{x}}\frac{dx}{dy}+\frac{1}{2\sqrt{y}}=0$
or $\frac{dx}{dy}=-\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{y}-4}{\sqrt{y}}$ ........................................Since $\sqrt{x}+\sqrt{y}=4$
or ${\left[\frac{dx}{dy}\right]}_{y=1}=\frac{1-4}{1}=-3$