Question

calculate the derivative of y in respect to x x^(3)y+3xy^(3)=x+y

Answer 1

guess my answer was not sufficient. do you know how to use implicit diff?

Answer 2

no. the teacher covered it in class today...sort of.

Answer 3

\[x^ 3y+3xy^3=x+y\] think of
\[x^3f(x)+3xf^2(x)=x+f(x)\]

Answer 4

you need to use the product rule to find the derivative of
\[x^3f(x)\] it is
\[3x^2f(x)+x^3f'(x)\] then replacing f(x) by y you get
\[3x^2y+x^3y'\]

Answer 5

for
\[3xy^3\] you need both the product rule and the chain rule

Answer 6

the derivative of
\[3xf^3(x)\] is
\[3f^3(x)+9xf^2(x)f'(x)\] or the derivative of
\[3xy^3\] is
\[3y+9xy^2y'\]

Answer 7

In this case, implicit differentiation has to be used.

So, you are finding derivative of y respect to x

x^(3)y+3xy^(3)=x+y

First, Find the derivative of x^3 y by using product rule

Derivative of x^3 y is 3x^2 y + dyx^3 .

Then, find the derivative of 3xy^3 with the same rule

Derivative of 3xy^3 is 3y^3 + 3y^2 (dy) (3x) = 3y^3 + 9xy^2 (dy)

After that, find the derivative from the other side of equation and you get,

1 + dy.

Let's write down everything back,

3x^2 y + dyx^3 + 3y^3 + 9xy^2 (dy) = 1 + dy

Send all of the dy to one side,

dyx^3 + 9xy^2 (dy) - dy = 1- 3x^2 y - 3y^3

Factor out the dy 's,

dy ( x^3 + 9xy^2 - 1) = 1 - 3x^2 y - 3y^3

Then, divide both side with ( x^3 + 9xy^2 - 1)

So you have dy = [ 1- 3x^2 y - 3y^3 ] / [x^3 + 9xy^2 - 1]

Hope, above helps.. srry if it's too long ...heh