Question

Find the first derivative for
y * sqrt(x) + x * sqrt(y) = 9

Answer 1

\[y'\sqrt{x}+\frac{1}{2\sqrt{x}}y+\sqrt{y}+\frac{xy'}{2\sqrt{y}}=0\] is a start

Answer 2

now you get to solve that mess for
\[y'\]

Answer 3

I don't understand where the \[(xy')\div2\sqrt{y}\] came from.

Answer 4

\[[y \sqrt{x}]'=(y)'\sqrt{x}+y(\sqrt{x})'=y'\sqrt{x}+y \cdot \frac{1}{ 2 \sqrt{x}}\]

\[[x \sqrt{y}]'=(x)'\sqrt{y}+x(\sqrt{y})'=1 \sqrt{y}+x \frac{1}{2 \sqrt{y}} \cdot y'\]

Answer 5

\[(\sqrt{y})'=(y^\frac{1}{2})'=\frac{1}{2}y^{\frac{1}{2}-1}y'=\frac{1}{2}y^\frac{-1}{2}y'=\frac{1}{2} \frac{1}{y^\frac{1}{2}}y' \]

Answer 6

don't forget chain rule is derivative of outside times derivative of inside

Answer 7

Oh okay I get why.

Answer 8

Thanks.

Answer 9

np :)