Question

find dy/dx by implicit differentiation at the point (-1,1)
(x+y)^3=x^3+y^3
someone please help

Answer 1

\[\Large (x+y)^3 = x^3 + y^3\]
\[\Large \frac{d}{dx}\left[(x+y)^3\right] = \frac{d}{dx}\left[x^3 + y^3\right]\]
\[\Large 3(x+y)^2*\frac{d}{dx}\left[(x+y)\right] = 3x^2 + 3y^2*\frac{dy}{dx}\]
\[\Large 3(x+y)^2*\left(1+\frac{dy}{dx}\right) = 3x^2 + 3y^2*\frac{dy}{dx}\]
The goal from here is to isolate dy/dx. I'll let you take over.

Answer 2

To make things a bit easier, you can plug in (x,y) = (-1,1) after you get done with the differentiation.

Answer 3

Another approach can be,

Using Pascas' triange or just expension:
\(\color{#000000 }{ \displaystyle (x+y)^3=x^3+y^3+3x^2y+3y^2x }\)

So,
\(\color{#000000 }{ \displaystyle (x+y)^3=x^3+y^3 \quad\Longrightarrow\quad3x^2y+3y^2x=0 }\)

\(\color{#000000 }{ \displaystyle\Longrightarrow \quad x^2y=y^2x}\)

Then, differentiate (dy/dx)
(y has a chain of y' because it is a also a function of x)

\(\color{#000000 }{ \displaystyle y'x^2+2yx=2yy'x+y^2 }\)

Simpify,
\(\color{#000000 }{ \displaystyle y'x^2-2yy'x=y^2-2yx\quad\Longrightarrow\quad y'=\frac{y^2-2yx }{x^2-2yx} }\)

Answer 4

Then plug in your point (-1,1).
(-1 for x, and 1 for y)

Answer 5

Refer to the attachment from the Mathematica v9 computer program.