Question

If $\sqrt{x}+\sqrt{y}=5$, find the value of $\frac{dy}{dx}$ at $\left(4,9\right)$.

• A. $\frac{3}{2}$
• B. $-\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (B)

$-\frac{3}{2}$

Given:- $\sqrt{x}+\sqrt{y}=5\Rightarrow \sqrt{y}=5-\sqrt{x}$

On squaring both side of above equation, $y=25+x-10\sqrt{x}$

Now, differentiating above equation w.r.t. x, we get

$\frac{dy}{dx}=\frac{d}{dx}(x+25-10\sqrt{x})=1+0-10\times \frac{1}{2}\times \frac{1}{\sqrt{x}}=1-\frac{5}{\sqrt{x}}$

Thus$,\frac{dy}{dx}$ at $(4,9)=1-\frac{5}{\sqrt{4}}=1-\frac{5}{2}=\frac{-3}{2}$