Answer this question and show the steps. sqrt{x}+sqrt{y}=3, find the value of frac{dy}{dx} at the point (4,1)

Question
Decimals
asked 2020-12-24
Answer this question and show the steps. \(\displaystyle\sqrt{{{x}}}+\sqrt{{{y}}}={3}\), find the value of \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\) at the point (4,1)

Answers (1)

2020-12-25
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).
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