Implicit Diff Problem

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Jason76

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Post Edited 9/27/13

Solving for y'

\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

\(\displaystyle 6x^{2} + y(2x) + x^{2}(1)y' - y^{3}(1) + x(3y^{2})y' = 0\)

\(\displaystyle 6x^{2} + 2xy + x^{2}y' - y^{3} + 3xy^{2}y' = 0\)

\(\displaystyle 6x^{2} + 2xy - y^{3} + x^{2}y' + 3xy^{2}y' = 0\)

\(\displaystyle 6x^{2} + 2xy - y^{3} + y'(x^{2} + 3xy^{2}) = 0\)

\(\displaystyle y'(x^{2} + 3xy^{2}) = -6x^{2} - 2xy + y^{3}\)

\(\displaystyle y' = \dfrac{ -6x^{2} - 2xy + y^{3}}{x^{2} + 3xy^{2}}\)
 
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Your second step is incorrect. You have to use the power rule to differentiate x^2 * y. The derivative of that term is: 2xy + dy/dx * x^2 . You made the same mistake with another term. When performing implicit differentiation this way, do not treat y as a constant. It is a variable, just like x, and it can be differentiated according to the same rules.

A shortcut to implicit differentiation is to use the result of the Implicit Function Theorem. According to that, the implicit derivative of a function of x and y is -f'(x)/f'(y), where f'(x) is the derivative of f(x) with respect to x, and f'(y) is the derivative of f(y) with respect to y.

First move all the terms to the left side, = 0. Then differentiate each term that contains x, treating y as a constant, as the numerator. Negate the numerator. Differentiate each term that contains y, treating x as a constant, as the denominator.

(I don't know the Latex code.)

2x^3 + x^2 *y - xy^3 -3 = 0

- (6x^2 +2xy -y^3)/(x^2 - 3xy^2) = (-6x^2 -2xy + y^3)/x(x-3y^2)
 
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Jason76 said:
Solving for y'

\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

If f(x) = 3 → f'(x) = 0

\(\displaystyle f'(x) = 6x^{2} + 2xyy' - 3y^{2}y' = 0\)

\(\displaystyle f'(x) = 2xyy' - 3y^{2}y' - 3y^{2}y' = -6x^{2}\)

\(\displaystyle f'(x) = y'y(2x - 3y) = -6x^{2}\)

\(\displaystyle f'(x) = y'y = \dfrac{ -6x^{2}}{(2x - 3y)} \)

\(\displaystyle f'(x) = yy'(\dfrac{1}{y}) = \dfrac{ -6x^{2}}{(2x - 3y)} (\dfrac{1}{y}) \)

\(\displaystyle f'(x) = y' = \dfrac{ -6x^{2}}{(2x - 3y^{2})}\):confused:
Click to expand...
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\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

\(\displaystyle f'(x) = 2x^{3} + y(2x)y' + x^{2}(1)y' - y^{3}(1)y' + x(3y^{2})y' = 0\)

\(\displaystyle f'(x) = 2x^{3} + 2xyy' + x^{2}y' - y^{3}y' + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 2x^{3} + y'(2xy - + x^{2}y' - y^{3} + 3xy^{2}= 0\)

\(\displaystyle f'(x) = 2x^{3} + y'(2xy + x^{2}y' - y^{3} + 3xy^{2}) = x^{2}\)

\(\displaystyle f'(x) = y'(2xy + x^{2}y' - y^{3} + 3xy^{2}) = x^{2}\)

\(\displaystyle y' = \dfrac{x^{2}}{2xy + x^{2}y' - y^{3} + 3xy^{2}}\):confused:
 
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Jason76 said:
\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

\(\displaystyle f'(x) = 2x^{3} + y(2x)y' + x^{2}(1)y' - y^{3}(1)y' + x(3y^{2})y' = 0\) .............. Incorrect .................should be

\(\displaystyle 6x^{2} + y(2x) + x^{2}(1)y' - y^{3}(1) - x(3y^{2})y' = 0\)

:confused:
Click to expand...
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Solving for y'

\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

\(\displaystyle f'(x) = 2x^{3} + y(2x) + x^{2}(1)y' - y^{3}(1) + x(3y^{2})y' = 0\)

\(\displaystyle f'(x) = 2x^{3} + 2xy + x^{2}y' - y^{3} + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 2x^{3} + 2xy - y^{3} + x^{2}y' + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 2x^{3} + 2xy - y^{3} + y'(x^{2} + 3xy^{2}) = 0\)

\(\displaystyle f'(x) = y'(x^{2} + 3xy^{2}) = -2x^{3} - 2xy + y^{3}\)
 
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Jason76 said:
Solving for y'

\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

\(\displaystyle f'(x) = 2x^{3} + y(2x) + x^{2}(1)y' - y^{3}(1) + x(3y^{2})y' = 0\) DUDE, WRONG! SEE SKHAN'S POST ABOVE!

\(\displaystyle f'(x) = 2x^{3} + 2xy + x^{2}y' - y^{3} + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 2x^{3} + 2xy - y^{3} + x^{2}y' + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 2x^{3} + 2xy - y^{3} + y'(x^{2} + 3xy^{2}) = 0\)

\(\displaystyle f'(x) = y'(x^{2} + 3xy^{2}) = -2x^{3} - 2xy + y^{3}\)
Click to expand...

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Post Edited 9/27/13

Solving for y'

\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

\(\displaystyle f'(x) = 6x^{2} + y(2x) + x^{2}(1)y' - y^{3}(1) + x(3y^{2})y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy + x^{2}y' - y^{3} + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + x^{2}y' + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + y'(x^{2} + 3xy^{2}) = 0\)

\(\displaystyle f'(x) = y'(x^{2} + 3xy^{2}) = -6x^{2} - 2xy + y^{3}\)

\(\displaystyle y' = \dfrac{ -6x^{2} - 2xy + y^{3}}{x^{2} + 3xy^{2}}\)
 
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Jason76 said:
Post Edited 9/27/13

Solving for y'

\(\displaystyle f(x) = 2x^{3} + x^{2}y - xy^{3} = 3\)

\(\displaystyle f'(x) = 6x^{2} + y(2x) + x^{2}(1)y' - y^{3}(1) + x(3y^{2})y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy + x^{2}y' - y^{3} + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + x^{2}y' + 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + y'(x^{2} + 3xy^{2}) = 0\)

\(\displaystyle f'(x) = y'(x^{2} + 3xy^{2}) = -6x^{2} - 2xy + y^{3}\)

\(\displaystyle y' = \dfrac{ -6x^{2} - 2xy + y^{3}}{x^{2} + 3xy^{2}}\)
Click to expand...

I believe SKhan had a typo in his derivation:

\(\displaystyle f'(x) = 6x^{2} + y(2x) + x^{2}(1)y' - y^{3}(1) - x(3y^{2})y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy + x^{2}y' - y^{3} - 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + x^{2}y' - 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + y'(x^{2} - 3xy^{2}) = 0\)

\(\displaystyle f'(x) = y'(x^{2} - 3xy^{2}) = -6x^{2} - 2xy + y^{3}\)

\(\displaystyle y' = \dfrac{ -6x^{2} - 2xy + y^{3}}{x^{2} - 3xy^{2}}\)
 
srmichael said:
I believe SKhan had a typo in his derivation:

\(\displaystyle f'(x) = 6x^{2} + y(2x) + x^{2}(1)y' - y^{3}(1) - x(3y^{2})y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy + x^{2}y' - y^{3} - 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + x^{2}y' - 3xy^{2}y' = 0\)

\(\displaystyle f'(x) = 6x^{2} + 2xy - y^{3} + y'(x^{2} - 3xy^{2}) = 0\)

\(\displaystyle f'(x) = y'(x^{2} - 3xy^{2}) = -6x^{2} - 2xy + y^{3}\)

\(\displaystyle y' = \dfrac{ -6x^{2} - 2xy + y^{3}}{x^{2} - 3xy^{2}}\)
Click to expand...

No, it was my mistake. I did some editing on some posts.
 
Sorry, but my online homework still says it's wrong. I wonder what is the problem?

Oh, wait. Never mind. I just typed it in wrong. :)
 
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