Q101 / 180 · KCET 2019 · Mathematics
If \( \sqrt[3]{y} \sqrt{x}=\sqrt[6]{(x+y)^{5}} \), then \( \frac{d y}{d x}= \)
- A\( x+y \)
- Correct answerB\( \frac{y}{x} \)
- C\( \frac{x}{y} \)
- D\( x-y \)
Solution
(B)
\[ \begin{array}{l} \sqrt[3]{y} \sqrt{x}=\sqrt[6]{(x+y)^{5}} \\ \Rightarrow y^{2} x^{3}=(x+y)^{5} \\ \Rightarrow y^{2} x^{3}=y^{5}\left(\frac{x}{y}+1\right)^{5} \\ \Rightarrow\left(\frac{x}{y}\right)^{3}=\left(\frac{x}{y}+1\right)^{5} \\ \Rightarrow \frac{d y}{d x}=\frac{y}{x} \\ f\left(\frac{x}{y}\right)=c \Rightarrow \frac{d y}{d x}=\frac{y}{x} \end{array} \]
\[ \begin{array}{l} \sqrt[3]{y} \sqrt{x}=\sqrt[6]{(x+y)^{5}} \\ \Rightarrow y^{2} x^{3}=(x+y)^{5} \\ \Rightarrow y^{2} x^{3}=y^{5}\left(\frac{x}{y}+1\right)^{5} \\ \Rightarrow\left(\frac{x}{y}\right)^{3}=\left(\frac{x}{y}+1\right)^{5} \\ \Rightarrow \frac{d y}{d x}=\frac{y}{x} \\ f\left(\frac{x}{y}\right)=c \Rightarrow \frac{d y}{d x}=\frac{y}{x} \end{array} \]
Hence, the correct answer is option: 'B'.