Question

If $\sqrt{x} \sqrt[3]{y} = (x + y)^n$ and $x\frac{dy}{dx} - y = 0$, then $n =$

• A. 1
• B. $\frac{6}{5}$
• C. $\frac{5}{6}$
• D. $\frac{4}{9}$

Hint:

Memorize this highly useful shortcut result for competitive exams: If $x^m \cdot y^n = (x+y)^{m+n}$, then it is guaranteed that $\frac{dy}{dx} = \frac{y}{x}$. Recognizing this pattern instantly solves the problem without any differentiation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given the relation \[ \sqrt{x}\,\sqrt[3]{y} = (x + y)^n \] and the differential condition \[ x\frac{dy}{dx} - y = 0 \;\Rightarrow\; \frac{dy}{dx} = \frac{y}{x}. \] We need to find the value of $n$.
Step 2: Rewrite the Equation:
Express the given equation in exponent form: \[ x^{\frac{1}{2}} \cdot y^{\frac{1}{3}} = (x + y)^n \]
Step 3: Logarithmic Differentiation:
Taking logarithm on both sides: \[ \frac{1}{2}\ln x + \frac{1}{3}\ln y = n \ln(x+y) \] Differentiate both sides w.r.t.\ $x$: \[ \frac{1}{2x} + \frac{1}{3y}\frac{dy}{dx} = n \cdot \frac{1}{x+y}\left(1 + \frac{dy}{dx}\right) \]
Step 4: Substitute Given Condition:
Using $\displaystyle \frac{dy}{dx} = \frac{y}{x}$: \[ \frac{1}{2x} + \frac{1}{3y}\left(\frac{y}{x}\right) = \frac{n}{x+y}\left(1 + \frac{y}{x}\right) \] \[ \frac{1}{2x} + \frac{1}{3x} = \frac{n}{x+y} \cdot \frac{x+y}{x} \] \[ \frac{3+2}{6x} = \frac{n}{x} \] \[ \frac{5}{6x} = \frac{n}{x} \]
Step 5: Solve for $n$:
Multiplying both sides by $x$: \[ n = \frac{5}{6} \]
Step 6: Final Answer:
\[ \boxed{\frac{5}{6}} \]