Implicit Differentiation of $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6$

Question

Find $\frac{dy}{dx}$ when $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6$.

Answer 1

This was originally an answer for @Sanchayan Dutta but for some reason it became closed. Anyway, here's an answer.

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You can implicitly differentiate as is, but that's no fun! So let's rearrange the equation so we only have to perform one product rule differentiation:

$$\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6. $$

Just going to use the notation $\sqrt{z} = z^{1/2}$ for convenience: $$ x^{1/2} y^{-1/2} + x^{-1/2}y^{1/2} = 6 \\ x^{-1/2} (xy^{-1/2} + y^{1/2}) = 6 \\ x^{-1/2} y^{-1/2} ( x + y) = 6 \\ \frac{x+y}{x^{1/2}y^{1/2}} = 6 \\ x+y = 6x^{1/2}y^{1/2}.$$

This looks much less intimidating than the original expression. Okay, now we implicitly differentiate both sides with respect to $x$. Note how the product rule is applied: $$\frac{d}{dx} [x+y\big] = \frac{d}{dx} [6x^{1/2}y^{1/2}] \\ \frac{dx}{dx} + \frac{dy}{dx} = 6\left(x^{1/2}(1/2)y^{-1/2}\frac{dy}{dx} + (1/2)x^{-1/2} \frac{dx}{dx} y^{1/2}\right).$$

Now let's cancel out those pesky $\frac{dx}{dx}$'s and change them to $1$'s, and let the dust settle by cleaning up the fractions:

$$1+\frac{dy}{dx} = 3 x^{1/2}y^{-1/2} \frac{dy}{dx} + 3x^{-1/2}y^{1/2}.$$

We're now in a position to simply isolate $\frac{dy}{dx}$:

$$1+\frac{dy}{dx} = 3 x^{1/2}y^{-1/2} \frac{dy}{dx} + 3x^{-1/2}y^{1/2} \\ \frac{dy}{dx} - 3 x^{1/2}y^{-1/2} \frac{dy}{dx} = 3x^{-1/2}y^{1/2} -1 \\ \frac{dy}{dx} (1 - 3 x^{1/2}y^{-1/2}) = 3x^{-1/2}y^{1/2} -1 \\ \frac{dy}{dx} = \frac{ 3x^{-1/2}y^{1/2} -1}{ 1 - 3 x^{1/2}y^{-1/2} }.$$ Multiplying both the numerator and the denominator of the right-hand-side fraction by $\sqrt{xy} = x^{1/2}y^{1/2}$ gives:

$$ \frac{dy}{dx} = \frac{ (3x^{-1/2}y^{1/2} -1)(x^{1/2}y^{1/2})}{ (1 - 3 x^{1/2}y^{-1/2})(x^{1/2}y^{1/2}) } \\ \frac{dy}{dx} = \frac{ 3y -\sqrt{xy}}{ \sqrt{xy} - 3 x}. $$ It might appear we are done, but recall that $x+y = 6\sqrt{xy}$. So replacing the $\sqrt{xy}$ with a linear term of $\frac{1}{6} (x+y)$ gives $$ \frac{dy}{dx} = \frac{ 3y -\frac{1}{6}x - \frac{1}{6}y}{ \frac{1}{6}x + \frac{1}{6}y - 3 x} \\ \frac{dy}{dx} = \frac{ \frac{17}{6}y-\frac{1}{6}x }{ + \frac{1}{6}y - \frac{17}{6}x} \\ \frac{dy}{dx} = \frac{17y - x}{y - 17x}$$

which is probably as elegant as it's going to get.

Answer 2

We can multiply by $\sqrt{y/x}$ on both sides to get $1+y/x=6\sqrt{y/x}$. Let $v=\sqrt{y/x}$, which leaves us with $v^2-6v+1=0$. If we solve this quadratic, we are left with $$v=\frac{6\pm\sqrt{32}}{2}=3\pm 2\sqrt{2}.$$ Then $y=(3\pm2\sqrt{2})^2x.$ Thus, $$\frac{dy}{dx}=(3\pm2\sqrt{2})^2.$$