# implicit differentiation

• Oct 16th 2012, 01:13 PM
kingsolomonsgrave
implicit differentiation
I have a question about implicit differentiation.

In the case of differentiating the expression x^3+y^3=6xy the right hand side becomes 6xy'+6y where y is regarded as a function of x.

using the product rule $\displaystyle f'g+g'f$

$\displaystyle f = 6x$

$\displaystyle g = y$

then

$\displaystyle (6x)' y + 6y'$

so I get $\displaystyle 6y+6y'$

which is wrong.

Why is the right hand side equal to $\displaystyle 6xy'+6y$?

I (think) I know the key is that y is a function of x but somehow its not clicking as to why the answer is as it is.
• Oct 16th 2012, 01:52 PM
TheEmptySet
Re: implicit differentiation
Quote:

Originally Posted by kingsolomonsgrave
I have a question about implicit differentiation.

In the case of differentiating the expression x^3+y^3=6xy the right hand side becomes 6xy'+6y where y is regarded as a function of x.

using the product rule $\displaystyle f'g+g'f$

$\displaystyle f = 6x$

$\displaystyle g = y$

then

$\displaystyle (6x)' y + 6y'$

so I get $\displaystyle 6y+6y'$

which is wrong.

Why is the right hand side equal to $\displaystyle 6xy'+6y$?

I (think) I know the key is that y is a function of x but somehow its not clicking as to why the answer is as it is.

You differentated f is both parts of your product rule

It should be

$\displaystyle f(x)=6x \implies f'(x)=6 \quad g(x)=y \implies g'(x)=y'$

So

$\displaystyle \frac{d}{dx}(fg)=f'g+fg'=6y+(6x)y'$
• Oct 17th 2012, 06:01 AM
kingsolomonsgrave
Re: implicit differentiation
thank you! I wrote my prof and he just wrote back - it should be 6y+6xy'

I was like....yes, we know this, it's in the text book....talk about not helpful.

This forum (and the helpful math experts that populate it) save my life.

thanks again