Find dy/dx: (x + y)^2 = 4

Monday, August 20, 2012

Find dy/dx: (x + y)^2 = 4

Hello!

Let's differentiate the given equation:

$\displaystyle{\left({{\left({x}+{y}{\left({x}\right)}\right)}}^{{2}}\right)}'={4}',$

$\displaystyle{2}{\left({x}+{y}{\left({x}\right)}\right)}{\left({x}+{y}{\left({x}\right)}\right)}'={0},$

$\displaystyle{2}{\left({x}+{y}\right)}{\left({1}+{y}'\right)}={0}.$

We know that $\displaystyle{{\left({x}+{y}\right)}}^{{2}}={4},$ therefore $\displaystyle{x}+{y}=\pm{2}$ and it is never zero. Thus $\displaystyle{1}+{y}'{\left({x}\right)}={0}$ and $\displaystyle{y}'{\left({x}\right)}=-{1}.$ This is the answer (dy/dx and y'(x) are the different notations of the same notion).

Ninenotes

Monday, August 20, 2012

Find dy/dx: (x + y)^2 = 4

No comments:

Post a Comment

find square roots of -1+2i