In Exercises 5 – 6, evaluate fx(x, y) and fy(x, y) at the indicated point.
5. f(x, y) = x 2 y − x + 2y + 3 at (1, 2)
6. f(x, y) = x 3 − 3x + y 2 − 6y at (−1, 3)
\[ Given f(x,y)=x2y−x+2y+3 \]
To find \(fx(x,y),\) we take the partial derivative of f with respect to x: \[fx(x,y)=∂x∂(x2y−x+2y+3)=2xy−1=2xy−1\]
To find \(fy(x,y),\) we take the partial derivative of f with respect to y:
\(fy(x,y)=∂y∂(x2y−x+2y+3)=x2+2\)
Then we evaluate these derivatives at the point (1,2)
\(fx(1, 2) = 2(1)(2) - 1 = 4 - 1 = 3\)
\(y(1, 2) = 1^2 + 2 = 1 + 2 = 3\)
\(So, fx(1, 2) = 3\)
and \(fy(1, 2) = 3\)
Given \(f(x,y)=x3−3x+y2−6y\)
To find $fx(x,y), $ we take the partial derivative of f with respect to x:
\(fx(x,y)=∂x∂(x3−3x+y2−6y)\)
\(=3x2−3\)
To find \(fy(x,y),\) we ake the partial derivative of f with respect to y:
\(fy(x,y)=∂y∂(x3−3x+y2−6y)\)
\(=2y−6\)
Then, we evaluate these derivatives at the point (−1,3):
\(fx(−1,3)=3(−1)2−3=3−3=0\)
\(fy(−1,3)=2(3)−6=6−6=0\)
So, \(fx(−1,3)=0\)
and \(fy(−1,3)=0.\)