- Thread starter
- #1

#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

\(\displaystyle (y-x)y'=y-x+8\) where \(\displaystyle y=x+4\sqrt{x+2}\)

So the derivative is \(\displaystyle y'=1+\frac{2}{\sqrt{x+2}}\)

and the LHS becomes \(\displaystyle (y-x)(1+\frac{2}{\sqrt{x+2}})=

y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=\)

\(\displaystyle x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}\)

and the RHS becomes \(\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8\)

which isn't equal but the answer key seems to think it is because it gives an interval of definition.