Krish Schmitt

2022-10-08

Using implicit differentiation, find $\frac{dy}{dx}$
${x}^{3}{y}^{6}=\left(x+y{\right)}^{9}$
My Attempt:
${x}^{3}{y}^{6}=\left(x+y{\right)}^{9}$
Differentiating both sides with respect to $x$, we get:
$\frac{d}{dx}\left({x}^{3}{y}^{6}\right)=\frac{d}{dx}\left(\left(x+y{\right)}^{9}\right)$
${x}^{3}.\frac{d}{dx}{y}^{6}+{y}^{6}.\frac{d}{dx}{x}^{3}=9\left(x+y{\right)}^{8}\frac{d}{dx}\left(x+y\right)$
$6{x}^{3}{y}^{5}.\frac{dy}{dx}+3{x}^{2}{y}^{6}=9\left(x+y{\right)}^{8}\left(1+\frac{dy}{dx}\right)$
$\frac{dy}{dx}=\frac{9\left(x+y{\right)}^{8}-3{x}^{2}y}{6{x}^{3}{y}^{5}-9\left(x+y{\right)}^{8}}$
$\frac{dy}{dx}=\frac{y}{x}$

Houston Ellis

I fixed a typo in your question, and with that, your implicit differentiation is perfectly correct. However, it looks difficult (as in, I cannot see how) to blackuce it from that to the expected simple answer.

However, if you're dealing only with the real domain it makes things a lot easier to take cube roots right at the outset. This simplifies things greatly.
${x}^{3}{y}^{6}=\left(x+y{\right)}^{9}$
For $x,y\in \mathbb{R}$,
$x{y}^{2}=\left(x+y{\right)}^{3}$
Differentiating implictly wrt $x$,
${y}^{2}+2y{y}^{\prime }x=3\left(x+y{\right)}^{2}\left(1+{y}^{\prime }\right)$
${y}^{\prime }\left(2yx-3\left(x+y{\right)}^{2}\right)=3\left(x+y{\right)}^{2}-{y}^{2}$
${y}^{\prime }=\frac{3\left(x+y{\right)}^{2}-{y}^{2}}{2yx-3\left(x+y{\right)}^{2}}$
Use the initial relationship given on the RHS,
${y}^{\prime }=\frac{\frac{3\left(x+y{\right)}^{3}}{x+y}-{y}^{2}}{2yx-\frac{3\left(x+y{\right)}^{3}}{x+y}}$
${y}^{\prime }=\frac{\frac{3x{y}^{2}}{x+y}-{y}^{2}}{2yx-\frac{3x{y}^{2}}{x+y}}$
The rest is just algebra,
${y}^{\prime }=\frac{3x{y}^{2}-{y}^{2}\left(x+y\right)}{2yx\left(x+y\right)-3x{y}^{2}}$
${y}^{\prime }=\frac{3x{y}^{2}-x{y}^{2}-{y}^{3}}{2y{x}^{2}+2{y}^{2}x-3x{y}^{2}}$
${y}^{\prime }=\frac{2x{y}^{2}-{y}^{3}}{2y{x}^{2}-x{y}^{2}}$
${y}^{\prime }=\frac{{y}^{2}\left(2x-y\right)}{yx\left(2x-y\right)}$
${y}^{\prime }=\frac{y}{x}$
as requiblack, et voilà!

EDIT: And just to flesh out hint involving logarithmic differentiation:
$3\mathrm{log}x+6\mathrm{log}y=9\mathrm{log}\left(x+y\right)$
$\frac{3}{x}+\frac{6}{y}{y}^{\prime }=\frac{9}{x+y}\left(1+{y}^{\prime }\right)$
${y}^{\prime }\left(\frac{6}{y}-\frac{9}{x+y}\right)=\frac{9}{x+y}-\frac{3}{x}$
${y}^{\prime }\left(\frac{6x-3y}{y\left(x+y\right)}\right)=\frac{6x-3y}{x\left(x+y\right)}$
${y}^{\prime }=\frac{y}{x}$
A decidedly much quicker route to the answer.

spatularificw2

For
$\left(x+y{\right)}^{m+n}={x}^{m}{y}^{n}$
Take logarithm wrt base $e,$
$\left(m+n\right)\mathrm{ln}\left(x+y\right)=m\mathrm{ln}x+n\mathrm{ln}y$
Now differentiate wrt $x$

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