Question

If $\sqrt[3]{y}\sqrt{x}=\sqrt[6]{{\left(x+y\right)}^5},$ then $\frac{dy}{dx}=$

• A. x - y
• B. $\frac{x}{y}$
• C. $\frac{y}{x}$
• D. x + y

Answer 1

Answer: (C)

$\sqrt[3]{y}\sqrt{x}=\sqrt[6]{{\left(x+y\right)}^5}$
$\Rightarrow y^2{x}^3={\left(x+y\right)}^5$
$\Rightarrow y^2{x}^3=y^5{\left(\frac{x}{y}+1\right)}^5$
$\Rightarrow {\left(\frac{x}{y}\right)}^3={\left(\frac{x}{y}+1\right)}^5$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$
$f\left(\frac{x}{y}\right)=c$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$