KCET · Maths · Differentiation
If \( \sqrt[3]{y} \sqrt{x}=\sqrt[6]{(x+y)^{5}} \), then \( \frac{d y}{d x}= \)
- A \( x+y \)
- B \( \frac{y}{x} \)
- C \( \frac{x}{y} \)
- D \( x-y \)
Answer & Solution
Correct Answer
(B) \( \frac{y}{x} \)
Step-by-step Solution
Detailed explanation
(B)
\[
\begin{array}{l}
\sqrt[3]{y} \sqrt{x}=\sqrt[6]{(x+y)^{5}} \\
\Rightarrow y^{2} x^{3}=(x+y)^{5} \\
\Rightarrow y^{2} x^{3}=y^{5}\left(\frac{x}{y}+1\right)^{5} \\
\Rightarrow\left(\frac{x}{y}\right)^{3}=\left(\frac{x}{y}+1\right)^{5} \\
\Rightarrow \frac{d y}{d x}=\frac{y}{x} \\
f\left(\frac{x}{y}\right)=c \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{array}
\]
\[
\begin{array}{l}
\sqrt[3]{y} \sqrt{x}=\sqrt[6]{(x+y)^{5}} \\
\Rightarrow y^{2} x^{3}=(x+y)^{5} \\
\Rightarrow y^{2} x^{3}=y^{5}\left(\frac{x}{y}+1\right)^{5} \\
\Rightarrow\left(\frac{x}{y}\right)^{3}=\left(\frac{x}{y}+1\right)^{5} \\
\Rightarrow \frac{d y}{d x}=\frac{y}{x} \\
f\left(\frac{x}{y}\right)=c \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{array}
\]
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