# How to find dy/dx from x^3y^6=(x+y)^9 using implicit differentiation?

How to find
$dy/dx$
from
${x}^{3}{y}^{6}=\left(x+y{\right)}^{9}$
using implicit differentiation? I tried solving but I ended up with solution that does not agree with my textbook answer. How can I get
$dy/dx=y/x$
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rmercierm7
The answer does appear to be correct. Here's how to go about it:

1) First, use implicit differentiation to get
$3{x}^{2}{y}^{6}+6{x}^{3}{y}^{5}\frac{dy}{dx}=9\left(x+y{\right)}^{8}\left(1+\frac{dy}{dx}\right).$
2) Solve for $\frac{dy}{dx}$ to get
$\frac{dy}{dx}=\frac{9\left(x+y{\right)}^{8}-3{x}^{2}{y}^{6}}{6{x}^{3}{y}^{5}-9\left(x+y{\right)}^{8}}.$
3) Multiply the numerator and denominator of the RHS by $\left(x+y\right)$ to get
$\frac{dy}{dx}=\frac{9\left(x+y{\right)}^{9}-3{x}^{2}{y}^{6}\left(x+y\right)}{6{x}^{3}{y}^{5}\left(x+y\right)-9\left(x+y{\right)}^{9}}.$
4) Using the original formula for $\left(x+y{\right)}^{9}$, this becomes
$\frac{dy}{dx}=\frac{9{x}^{3}{y}^{6}-3{x}^{2}{y}^{6}\left(x+y\right)}{6{x}^{3}{y}^{5}\left(x+y\right)-9{x}^{3}{y}^{6}}.$
5) By simplifying, we get
$\frac{dy}{dx}=\frac{6{x}^{3}{y}^{6}-3{x}^{2}{y}^{7}}{6{x}^{4}{y}^{5}-3{x}^{3}{y}^{6}}=\frac{2xy-{y}^{2}}{2{x}^{2}-xy}.$
6) By factoring common factors, we get
$\frac{dy}{dx}=\frac{y\left(2x-y\right)}{x\left(2x-y\right)}=\frac{y}{x}.$
###### Did you like this example?
babuliaam
First simplify your equation by taking the cube root of each side to get
$x{y}^{2}=\left(x+y{\right)}^{3}$
Then take the derivative of both sides and do the usual process to find
$\frac{dy}{dx}=\frac{3\left(x+y{\right)}^{2}-{y}^{2}}{2xy-3\left(x+y{\right)}^{2}}$
Then multiplying numerator and denominator by $x+y$,
$\frac{dy}{dx}=\frac{3\left(x+y{\right)}^{3}-{y}^{2}\left(x+y\right)}{2xy\left(x+y\right)-3\left(x+y{\right)}^{2}}$
$=\frac{3x{y}^{2}-x{y}^{2}-{y}^{3}}{2{x}^{2}y+2x{y}^{2}-3x{y}^{2}}$
$=\frac{2xy-{y}^{3}}{2{x}^{2}y-x{y}^{2}}$
$=\frac{{y}^{2}\left(2-y\right)}{xy\left(2-y\right)}$
$=\frac{y}{x}$