Question

If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}} = 6\), then \(\frac{dy}{dx} =\)

• A. \(\frac{x+17y}{17x-y}\)
• B. \(\frac{x-17y}{17x-y}\)
• C. \(\frac{x-17y}{17x+y}\)
• D. \(\frac{x+17y}{17x+y}\)

Hint:

Let t = sqrt(x/y) and turn the equation into a relation between x and y without roots.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Put \(t=\sqrt{x/y}\). The equation becomes \(t+\dfrac1t=6\).

Step 2: Remove the roots
Square both sides:
\[ \frac xy+\frac yx+2=36\Rightarrow\frac{x^2+y^2}{xy}=34 \]
\[ x^2+y^2=34xy \]

Step 3: Differentiate
\[ 2x+2y\frac{dy}{dx}=34y+34x\frac{dy}{dx} \]
\[ \frac{dy}{dx}(2y-34x)=34y-2x \]
\[ \frac{dy}{dx}=\frac{17y-x}{y-17x}=\frac{x-17y}{17x-y} \]
This is option (B).

Final Answer:
After squaring, x squared + y squared = 34xy, and differentiating gives (x - 17y)/(17x - y), option (B). \[ \boxed{\frac{x-17y}{17x-y}} \]