Quod Erat Demonstrandum


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}


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.

2. How about when students start with

= \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.


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