# Calculus Problem

1. Mar 3, 2005

### erik05

If you have an equation such as this:

$$x^3-6xy+6=0$$

Would you solve this equation with implicit differentiation or could you solve for y to get $$y= \frac {x^3+6}{6x}$$ and then use the Quotient Rule to solve for its derivative? Just need some clarification....thanks in advance.

2. Mar 3, 2005

### dextercioby

If you don't know the theorem of implicit functions,then it's reccomendable to do the explicitation...In your case,it's fortunately unique...

Proceed with the differentiation.It's much useful to divide the numerator through the denominator...No need for quotient rule.

Daniel.

3. Mar 3, 2005

### erik05

But you still could use the Quotient Rule I'm assuming? This question was on a calculus test I had today and I worked it out both ways,with implicit and solving it explicitly and I got different answers. Implicit I got:

$$\frac {6y-3x^2}{6x}$$

Explicit,using the Quotient Rule I got:

$$\frac {x^3-3}{3x^2}$$

Or perhaps, both these answers are wrong and I should be hoping for pity marks right now. *sigh* I guess I'm just looking for some reassurance.

4. Mar 3, 2005

### dextercioby

The second answer is the correct one...

Daniel.

5. Mar 3, 2005

### HallsofIvy

Staff Emeritus
Just to clarify: implicit differentiation works just as well as direct differentiation- but you have the sign wrong in your implicit differentiation.

If x3- 6xy+ 6= 0 then, differentiating both sides with respect to x:

3x2- 6y- 6xy'= 0 so 6xy'= 3x2- 6y and

$$y'= \frac{3x^2- 6y}{6x}$$

Since $y= \frac{x^3+ 6}{6x}$ , $6y= \frac{x^3+ 6}{x}$,
$3x^2- 6y= \frac{3x^4- x^3- 6}{x}= \frac{2x^3- 6}{x}$ and so
$$y'= \frac{2x^3- 6}{6x}= \frac{x^3- 3}{3x^2}$$
exactly what you got by using direct differentiation.

Last edited: Mar 4, 2005
6. Mar 3, 2005

### erik05

Thanks for the clarification