Krish Schmitt

2022-10-08

Using implicit differentiation, find $\frac{dy}{dx}$

${x}^{3}{y}^{6}=(x+y{)}^{9}$

My Attempt:

${x}^{3}{y}^{6}=(x+y{)}^{9}$

Differentiating both sides with respect to $x$, we get:

$\frac{d}{dx}}({x}^{3}{y}^{6})={\displaystyle \frac{d}{dx}}((x+y{)}^{9})$

${x}^{3}.{\displaystyle \frac{d}{dx}}{y}^{6}+{y}^{6}.{\displaystyle \frac{d}{dx}}{x}^{3}=9(x+y{)}^{8}{\displaystyle \frac{d}{dx}}(x+y)$

$6{x}^{3}{y}^{5}.{\displaystyle \frac{dy}{dx}}+3{x}^{2}{y}^{6}=9(x+y{)}^{8}(1+{\displaystyle \frac{dy}{dx}})$

$\frac{dy}{dx}}={\displaystyle \frac{9(x+y{)}^{8}-3{x}^{2}y}{6{x}^{3}{y}^{5}-9(x+y{)}^{8}}$

But the answer is

$\frac{dy}{dx}}={\displaystyle \frac{y}{x}$

${x}^{3}{y}^{6}=(x+y{)}^{9}$

My Attempt:

${x}^{3}{y}^{6}=(x+y{)}^{9}$

Differentiating both sides with respect to $x$, we get:

$\frac{d}{dx}}({x}^{3}{y}^{6})={\displaystyle \frac{d}{dx}}((x+y{)}^{9})$

${x}^{3}.{\displaystyle \frac{d}{dx}}{y}^{6}+{y}^{6}.{\displaystyle \frac{d}{dx}}{x}^{3}=9(x+y{)}^{8}{\displaystyle \frac{d}{dx}}(x+y)$

$6{x}^{3}{y}^{5}.{\displaystyle \frac{dy}{dx}}+3{x}^{2}{y}^{6}=9(x+y{)}^{8}(1+{\displaystyle \frac{dy}{dx}})$

$\frac{dy}{dx}}={\displaystyle \frac{9(x+y{)}^{8}-3{x}^{2}y}{6{x}^{3}{y}^{5}-9(x+y{)}^{8}}$

But the answer is

$\frac{dy}{dx}}={\displaystyle \frac{y}{x}$

Houston Ellis

Beginner2022-10-09Added 4 answers

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.

${x}^{3}{y}^{6}=(x+y{)}^{9}$

For $x,y\in \mathbb{R}$,

$x{y}^{2}=(x+y{)}^{3}$

Differentiating implictly wrt $x$,

${y}^{2}+2y{y}^{\prime}x=3(x+y{)}^{2}(1+{y}^{\prime})$

${y}^{\prime}(2yx-3(x+y{)}^{2})=3(x+y{)}^{2}-{y}^{2}$

$y}^{\prime}=\frac{3(x+y{)}^{2}-{y}^{2}}{2yx-3(x+y{)}^{2}$

Use the initial relationship given on the RHS,

$y}^{\prime}=\frac{\frac{3(x+y{)}^{3}}{x+y}-{y}^{2}}{2yx-\frac{3(x+y{)}^{3}}{x+y}$

$y}^{\prime}=\frac{\frac{3x{y}^{2}}{x+y}-{y}^{2}}{2yx-\frac{3x{y}^{2}}{x+y}$

The rest is just algebra,

$y}^{\prime}=\frac{3x{y}^{2}-{y}^{2}(x+y)}{2yx(x+y)-3x{y}^{2}$

$y}^{\prime}=\frac{3x{y}^{2}-x{y}^{2}-{y}^{3}}{2y{x}^{2}+2{y}^{2}x-3x{y}^{2}$

$y}^{\prime}=\frac{2x{y}^{2}-{y}^{3}}{2y{x}^{2}-x{y}^{2}$

$y}^{\prime}=\frac{{y}^{2}(2x-y)}{yx(2x-y)$

$y}^{\prime}=\frac{y}{x$

as requiblack, et voilà!

EDIT: And just to flesh out hint involving logarithmic differentiation:

$3\mathrm{log}x+6\mathrm{log}y=9\mathrm{log}(x+y)$

$\frac{3}{x}+\frac{6}{y}{y}^{\prime}=\frac{9}{x+y}(1+{y}^{\prime})$

$y}^{\prime}(\frac{6}{y}-\frac{9}{x+y})=\frac{9}{x+y}-\frac{3}{x$

$y}^{\prime}(\frac{6x-3y}{y(x+y)})=\frac{6x-3y}{x(x+y)$

$y}^{\prime}=\frac{y}{x$

A decidedly much quicker route to the 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.

${x}^{3}{y}^{6}=(x+y{)}^{9}$

For $x,y\in \mathbb{R}$,

$x{y}^{2}=(x+y{)}^{3}$

Differentiating implictly wrt $x$,

${y}^{2}+2y{y}^{\prime}x=3(x+y{)}^{2}(1+{y}^{\prime})$

${y}^{\prime}(2yx-3(x+y{)}^{2})=3(x+y{)}^{2}-{y}^{2}$

$y}^{\prime}=\frac{3(x+y{)}^{2}-{y}^{2}}{2yx-3(x+y{)}^{2}$

Use the initial relationship given on the RHS,

$y}^{\prime}=\frac{\frac{3(x+y{)}^{3}}{x+y}-{y}^{2}}{2yx-\frac{3(x+y{)}^{3}}{x+y}$

$y}^{\prime}=\frac{\frac{3x{y}^{2}}{x+y}-{y}^{2}}{2yx-\frac{3x{y}^{2}}{x+y}$

The rest is just algebra,

$y}^{\prime}=\frac{3x{y}^{2}-{y}^{2}(x+y)}{2yx(x+y)-3x{y}^{2}$

$y}^{\prime}=\frac{3x{y}^{2}-x{y}^{2}-{y}^{3}}{2y{x}^{2}+2{y}^{2}x-3x{y}^{2}$

$y}^{\prime}=\frac{2x{y}^{2}-{y}^{3}}{2y{x}^{2}-x{y}^{2}$

$y}^{\prime}=\frac{{y}^{2}(2x-y)}{yx(2x-y)$

$y}^{\prime}=\frac{y}{x$

as requiblack, et voilà!

EDIT: And just to flesh out hint involving logarithmic differentiation:

$3\mathrm{log}x+6\mathrm{log}y=9\mathrm{log}(x+y)$

$\frac{3}{x}+\frac{6}{y}{y}^{\prime}=\frac{9}{x+y}(1+{y}^{\prime})$

$y}^{\prime}(\frac{6}{y}-\frac{9}{x+y})=\frac{9}{x+y}-\frac{3}{x$

$y}^{\prime}(\frac{6x-3y}{y(x+y)})=\frac{6x-3y}{x(x+y)$

$y}^{\prime}=\frac{y}{x$

A decidedly much quicker route to the answer.

spatularificw2

Beginner2022-10-10Added 1 answers

For

$(x+y{)}^{m+n}={x}^{m}{y}^{n}$

Take logarithm wrt base $e,$

$(m+n)\mathrm{ln}(x+y)=m\mathrm{ln}x+n\mathrm{ln}y$

Now differentiate wrt $x$

$(x+y{)}^{m+n}={x}^{m}{y}^{n}$

Take logarithm wrt base $e,$

$(m+n)\mathrm{ln}(x+y)=m\mathrm{ln}x+n\mathrm{ln}y$

Now differentiate wrt $x$

What is the derivative of the work function?

How to use implicit differentiation to find $\frac{dy}{dx}$ given $3{x}^{2}+3{y}^{2}=2$?

How to differentiate $y=\mathrm{log}{x}^{2}$?

The solution of a differential equation y′′+3y′+2y=0 is of the form

A) ${c}_{1}{e}^{x}+{c}_{2}{e}^{2x}$

B) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{3x}$

C) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}$

D) ${c}_{1}{e}^{-2x}+{c}_{2}{2}^{-x}$How to find instantaneous velocity from a position vs. time graph?

How to implicitly differentiate $\sqrt{xy}=x-2y$?

What is 2xy differentiated implicitly?

How to find the sum of the infinite geometric series given $1+\frac{2}{3}+\frac{4}{9}+...$?

Look at this series: 1.5, 2.3, 3.1, 3.9, ... What number should come next?

A. 4.2

B. 4.4

C. 4.7

D. 5.1What is the derivative of $\frac{x+1}{y}$?

How to find the sum of the infinite geometric series 0.9 + 0.09 + 0.009 +…?

How to find the volume of a cone using an integral?

What is the surface area of the solid created by revolving $f\left(x\right)={e}^{2-x},x\in [1,2]$ around the x axis?

How to differentiate ${x}^{\frac{2}{3}}+{y}^{\frac{2}{3}}=4$?

The differential coefficient of $\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$.