We need to find the derivatives with respect to x and y separately and then evaluate them at the given point.
a. Partial derivative with respect to x \(f_x(x,y)\)
\(\frac{df}{dx} x^3 - 3x + y^2 -6y\)
Differentiate each term separately
\(\frac{df}{dx} x^3 = 3x^2\)
\(\frac{df}{dx} -3x = -3\)
\(\frac{df}{dx} y^2\) and \(\frac{df}{dx} -6y\) don’t have any x’s their derivatives with respect to x are 0
\(f_x(x,y) = 3x^2 - 3\)
b. Partial derivative with respect to y \(f_y(x,y)\)
$ x^3 - 3x + y^2 -6y$ Differentiate each term separately
\(\frac{df}{dy} x^3\) and \(\frac{df}{dy} -3x\) don’t have any y’s so their derivatives with respect to y are 0
\(\frac{df}{dy} y^2 = 2y\)
\(\frac{df}{dy} -6y = -6\)
\(f_y(x,y) = 2y - 6\)
c. Evaluating \(f_x(x,y)\) and \(f_y(x,y)\) at point (-1,3)
\(f_x(x,y) = 3x^2 - 3\) = \(3(-1)^2 - 3\) = 0
\(f_y(x,y) = 2y - 6\) = \(2(3) - 6\) = 0
a. Partial derivative with respect to x \(f_x(x,y)\)
\(\frac{df}{dx} ln(xy)\)
Using the chain rule we get…
\(\frac{df}{dx} ln(xy) = \frac{1}{xy} * y\)
\(f_x(x,y) = \frac{1}{x}\)
b. Partial derivative with respect to y \(f_y(x,y)\)
\(\frac{df}{dy} ln(xy)\)
Using the chain rule we get…
\(\frac{df}{dy} ln(xy) = \frac{1}{xy} * x\)
\(f_y(x,y) = \frac{1}{y}\)
c. Evaluating \(f_x(x,y)\) and \(f_y(x,y)\) at point (-2,-3)
\(f_x(x,y) = \frac{1}{x}\) = \(\frac{1}{(-2)}\) = \(-\frac{1}{2}\)
\(f_y(x,y) = \frac{1}{y}\) = \(\frac{1}{(-3)}\) = \(-\frac{1}{3}\)