Question

Evaluate \(f_{x}, f_{y}\), and \(f_{z}\) at the given point. \(f(x, y, z)=\frac{x y}{x+y+z}, \quad(3,1,-1)\)

Short answer

The partial derivative of \(f\) with respect to \(x\) evaluated at \((3,1,-1)\) is \(\frac{1}{9}\), with respect to \(y\) is \(\frac{2}{3}\), and with respect to \(z\) is \(-\frac{1}{3}\).

Step-by-step solution

Derive \( f_{x} \)

Firstly, solve for the partial derivative of f with respect to x. To do this, we need to apply the quotient rule, which states that the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared. Applying this rule to the function \(f\), we get:\[f_{x}=\frac{(y)(x+y+z)-(xy)(1+1+0)}{(x+y+z)^2}\]

Derive \( f_{y} \)

Secondly, we find the partial derivative of f with respect to y. Similarly, we use the quotient rule, and we get:\[f_{y}=\frac{(x)(x+y+z)-(xy)(0+1+0)}{(x+y+z)^2}\] Among the three derivatives which need to be calculated, this is the simplest one because the derivatives of \(x\) and \(z\) (in terms of \(y\)) are zero.

Derive \( f_{z} \)

Lastly, we find the partial derivative of f with respect to z. Again, we apply the quotient rule considering that the derivative of \(x\) and \(y\) with respect to \(z\) is zero. We find:\[f_{z}=\frac{-(xy)(0+0+1)}{(x+y+z)^2}\]

Evaluate the partial derivatives

Now, we substitute the coordinates of the point \((3,1,-1)\) into each of the partial derivatives:\[f_{x}=\frac{(1)(3+1+(-1))-(3*1)(1+1+0)}{(3+1+(-1))^2}= \frac{1}{9}\]\[f_{y}=\frac{(3)(3+1+(-1))-(3*1)(0+1+0)}{(3+1+(-1))^2}= \frac{2}{3}\]\[f_{z}=\frac{-(3*1)(0+0+1)}{(3+1+(-1))^2}= -\frac{1}{3}\]