Question

Find y'(x) by implicit differentiation from sqrt(x)+sqrt(y)=1

Answer 1

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Answer 2

Differentiate both sides with respect to x and rearrange to solve for dy/dx.\[\sqrt{x}+\sqrt{y}=1 \rightarrow \frac{ 1 }{ 2 }x^{-1/2}+\frac{ 1 }{ 2 }y^{-1/2}\left( \frac{ dy }{ dx } \right)=0 \rightarrow \frac{ 1 }{ 2 }y^{-1/2}\left( \frac{ dy }{ dx } \right)=-\frac{ 1 }{ 2 }x^{-1/2}\]\[\rightarrow \left( \frac{ dy }{ dx } \right)=-\frac{ 1 }{ 2 }x^{-1/2} \div \frac{ 1 }{ 2 }y^{-1/2} \rightarrow \frac{ dy }{ dx }=-\frac{ 1 }{ 2\sqrt{x} } \times 2\sqrt{y} \rightarrow \frac{ dy }{ dx }=-\frac{ \sqrt{y} }{ \sqrt{x} }\]

@shyllat