Question

Find the gradient of the function at the given point. $$ z=\ln \left(x^{2}-y\right), \quad(2,3) $$

Short answer

The gradient of the function at the point (2,3) is (2, -1).

Step-by-step solution

Finding Partial Derivative with respect to \(x\)

Differentiate the given function partially with respect to \(x\). Use the chain rule: \( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \)\[\frac{\partial z}{\partial x}= \frac{1}{x^{2}-y} \cdot 2x = \frac{2x}{x^{2}-y}\]

Finding Partial Derivative with respect to \(y\)

Next, differentiate the given function partially respect to \(y\). Here, the derivative is negative because the function in the logarithm is subtracting \(y\).\[\frac{\partial z}{\partial y}= -\frac{1}{x^{2}-y}\]

Substituting Point into Partial Derivatives

Substitute the given point \((2,3)\) into the two derivatives we got from Steps 1 and 2. \[\frac{\partial z}{\partial x} \Bigg| _{(2,3)} = \frac{2*2}{2^{2}-3} = 2\] \[\frac{\partial z}{\partial y} \Bigg| _{(2,3)} = -\frac{1}{2^{2}-3} = -1\]

Writing the Gradient

Finally, write the gradient as an ordered pair (or vector), with the derivative with respect to \(x\) as the first element and the derivative with respect to \(y\) as the second element. \[ \nabla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) = (2, -1) \]