# implicit differentiation

• Dec 14th 2009, 05:08 PM
Evan.Kimia
implicit differentiation
I have a quick question regarding implicit differentiation:
Heres the problem and its solution
.

http://img697.imageshack.us/img697/5...91214at904.png

What I dont understand is from the 1st step (First arrow) to the second, in particular why the x^2 turns into y^2 in 2xycos(y^2). The same goes on the right side. Why isnt it y'[2xycos(x^2)+sin(x^2)]=-sin(y^2)-2xycos(y^2)?

Also, on the right side there is a y' which seems to disappear?

• Dec 14th 2009, 05:09 PM
Prove It
Quote:

Originally Posted by Evan.Kimia
I have a quick question regarding implicit differentiation:
Heres the problem and its solution
.

http://img697.imageshack.us/img697/5...91214at904.png

What I dont understand is from the 1st step (First arrow) to the second, in particular why the x^2 turns into y^2 in 2xycos(y^2). The same goes on the right side. Why isnt it y'[2xycos(x^2)+sin(x^2)]=-sin(y^2)-2xycos(y^2)?

Also, on the right side there is a y' which seems to disappear?

They collected all the terms which had a \$\displaystyle y'\$ on one side of the equation and all the terms without on the other side.

This is so you can take out a common factor of \$\displaystyle y'\$ and thus solve for \$\displaystyle y'\$.
• Dec 14th 2009, 05:30 PM
Evan.Kimia
Quote:

Originally Posted by Prove It
They collected all the terms which had a \$\displaystyle y'\$ on one side of the equation and all the terms without on the other side.

This is so you can take out a common factor of \$\displaystyle y'\$ and thus solve for \$\displaystyle y'\$.

Ohhh. I remember now, thank you for clarifying.