# Quod Erat Demonstrandum

## 2009/02/24

### Minor point in differentiation

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 1:43 下午
Tags: ,

This is a simple question in differentiation.

Let $x^2 = \sqrt{y^6}$ for any real number $y$, determine $\frac{dy}{dx}$ at (1,-1).

Some students may give the following: $x^2 = \sqrt{y^6}$ $x^2 = y^3$ – – – – – – (*)

Then differentiate both sides with respect to $x$, yield $2x = 3y^2\frac{dy}{dx}$

Put $x = 1, y = -1$, yield $2(1) = 3(-1)^2\frac{dy}{dx}$ $\therefore \frac{dy}{dx} = \frac{2}{3}$

Wrong!

The correct answer should be $-\frac{2}{3}$.

The main problem appears in (*). $x^2 = y^3$ is true only when $y \ge 0$.

If $y < 0$, we will have the following $x^2 = -y^3$ – – – – – – (**)

As I mentioned in “16 踢 #2″, we should pay attention when taking square roots (say), that is $\sqrt{a^2}$ is NOT equal to $a$; $\sqrt{a^2}$ must be non-negative, hence we should write $\sqrt{a^2} = |a|$, that is, $\sqrt{a^2} = a$ when $a \ge 0$; $\sqrt{a^2} = -a$ when $a < 0$

Back to the original question, we would like to find the first derivative at (1,-1), and the $y$-coordinate is negative. The fact is $\sqrt{y^6} = -y^3$ (for $y < 0$)

and we should find the derivative out of the following relation $x^2 = -y^3$.

And the correct answer will be figured out.

Something more…

1. May be, it is better to start with $x^4 = y^6$, try. $\frac{d}{dx}\sqrt{y^6}$ $= \frac{d}{dx}(y^6)^{\frac{1}{2}}$ $= \frac{1}{2}(y^6)^{\frac{-1}{2}}(6y^5)\frac{dy}{dx}$

Give comment.

3. Let $x^2 = \sqrt{y^n}$ ( $n$ is an even positive integer) for any real number $y$, determine $\frac{dy}{dx}$ at (1,-1). [You may consider cases of $n = 4m$ and $n = 4m + 2$.]

Urm, here is something about “ $\sqrt{4} \ne 4^{\frac{1}{2}}$“, just read the old post.