Partial Derivative: find dy for f(x,y) = x^2y^3-x+y^2-50

[stars584]

Given f(x, y) = x^2y^3 - x + y^2 - 50

1. Find fy (partial deriv)

My answer: fy = (x^2)(3y^2) + (2y)

2. Find fyx

My answer: fyx = 2x3(y^2) = 6x(y^2)

3. Find the slope of the tangent line to the curve of intersection of the graph of f(x,y) and the plane x = 3 at the point where x - 3 and y = 2.

My answer:

. . .(3, 2, 23)
. . .(3^2)(2^3) - 3 + (2^2) - 50
. . .9*8 - 3 + 4 - 50 = 23

. . .(3^2)*(3*(2^2) + (2*2)
. . .9*12 + 4 = 112

4. Find a vector parallel to the tangent line in #3.

My answer: 0i + 3j + 112k

5. Find the parametric equations of the tangent line in #3.

My answer:
. . .(x - 3) = ot
. . .(y - 2) = 3t
. . .(z - 23) = 112t

Is this correct?
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Edited by stapel -- Reason for edit: formatting, content from second thread, etc

 

[pka]

It is better to use a standard form for your answers.
\(\displaystyle \L \begin{array}{l}
f(x,y) = x^2 y^3 - x + y^2 - 50 \\
f_y (x,y) = 3x^2 y^2 + 2y\quad \& \quad f_{yx} (x,y) = 6xy^2 \\
\end{array}\)