Step 1
Consider the given function
\(\displaystyle\sqrt{{{x}}}+\sqrt{{{y}}}={3}\)
Step 2
Differentiate implicitly with respect to x.
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left(\sqrt{{{x}}}+\sqrt{{{y}}}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({3}\right)}\)

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{\frac{{{1}}}{{{2}}}}}}\right)}+{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({y}^{{{\frac{{{1}}}{{{2}}}}}}\right)}={0}\)

\(\displaystyle{\frac{{{1}}}{{{2}}}}{x}^{{{\frac{{{1}}}{{{2}}}}-{1}}}+{\left({\frac{{{1}}}{{{2}}}}{y}^{{{\frac{{{1}}}{{{2}}}}-{1}}}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}\)

\(\displaystyle{\frac{{{1}}}{{{2}}}}{x}^{{-{\frac{{{1}}}{{{2}}}}}}+{\left({\frac{{{1}}}{{{2}}}}{y}^{{-{\frac{{{1}}}{{{2}}}}}}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}\)

\(\displaystyle{\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}+{\frac{{{1}}}{{{2}\sqrt{{{y}}}}}}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{\sqrt{{{y}}}}}{{\sqrt{{{x}}}}}}=-\sqrt{{{\frac{{{y}}}{{{x}}}}}}\) Step 3 Now, find the value of \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\) at the point (4, 1) \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}{\left({x},{y}\right)}={\left({4},{1}\right)}=-\sqrt{{{\frac{{{1}}}{{{4}}}}}}=-{\frac{{{1}}}{{{2}}}}\) The answer is \(\displaystyle-{\frac{{{1}}}{{{2}}}}\) or -0.5(in decimals).

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{\frac{{{1}}}{{{2}}}}}}\right)}+{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({y}^{{{\frac{{{1}}}{{{2}}}}}}\right)}={0}\)

\(\displaystyle{\frac{{{1}}}{{{2}}}}{x}^{{{\frac{{{1}}}{{{2}}}}-{1}}}+{\left({\frac{{{1}}}{{{2}}}}{y}^{{{\frac{{{1}}}{{{2}}}}-{1}}}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}\)

\(\displaystyle{\frac{{{1}}}{{{2}}}}{x}^{{-{\frac{{{1}}}{{{2}}}}}}+{\left({\frac{{{1}}}{{{2}}}}{y}^{{-{\frac{{{1}}}{{{2}}}}}}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}\)

\(\displaystyle{\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}+{\frac{{{1}}}{{{2}\sqrt{{{y}}}}}}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{\sqrt{{{y}}}}}{{\sqrt{{{x}}}}}}=-\sqrt{{{\frac{{{y}}}{{{x}}}}}}\) Step 3 Now, find the value of \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\) at the point (4, 1) \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}{\left({x},{y}\right)}={\left({4},{1}\right)}=-\sqrt{{{\frac{{{1}}}{{{4}}}}}}=-{\frac{{{1}}}{{{2}}}}\) The answer is \(\displaystyle-{\frac{{{1}}}{{{2}}}}\) or -0.5(in decimals).