# Quod Erat Demonstrandum

## 2019/05/05

### What’s wrong?

Filed under: Additional / Applied Mathematics,NSS — johnmayhk @ 7:05 下午
Tags: ,

Here is a basic level M2 question:

Given that $\sqrt{xy}=7+2y$, find $\frac{dy}{dx}$ at ($-\frac{1}{3}$,$-3$).

Student 1 gave

$\frac{1}{2\sqrt{xy}}(x\frac{dy}{dx}+y)=2\frac{dy}{dx}$

$\frac{1}{2}\sqrt{\frac{x}{y}}\frac{dy}{dx}+\frac{1}{2}\sqrt{\frac{y}{x}}=2\frac{dy}{dx}$

$\frac{dy}{dx}=\sqrt{\frac{y}{x}}\cdot\frac{1}{4-\sqrt{\frac{x}{y}}}$

Thus, at ($-\frac{1}{3}$,$-3$),

$\frac{dy}{dx}=\sqrt{\frac{-3}{-1/3}}\cdot\frac{1}{4-\sqrt{\frac{-1/3}{-3}}}=\frac{9}{11}$

Student 2 gave

$\sqrt{x}\cdot \frac{1}{2\sqrt{y}}\cdot \frac{dy}{dx}+\sqrt{y}\cdot \frac{1}{2\sqrt{x}}=2\frac{dy}{dx}$

$\frac{1}{2}\sqrt{\frac{x}{y}}\frac{dy}{dx}+\frac{1}{2}\sqrt{\frac{y}{x}}=2\frac{dy}{dx}$

Thus, at ($-\frac{1}{3}$,$-3$),

$\frac{1}{2}\sqrt{\frac{-1/3}{-3}}\frac{dy}{dx}+\frac{1}{2}\sqrt{\frac{-3}{-1/3}}=2\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{9}{11}$

The answer obtained above is wrong.

By using Desmos, the shape of the equation $\sqrt{xy}=7+2y$ can be easily seen below,

thus $\frac{dy}{dx}$ at ($-\frac{1}{3}$,$-3$) should be negative, in fact, the correct answer is $-\frac{9}{13}$.

Students, what’s wrong there?

1. Is it always true that $\frac{x}{\sqrt{xy}}=\sqrt{\frac{x}{y}}$ ?

2. Is it always true that $\sqrt{xy}=\sqrt{x}\sqrt{y}$ ?