Find the first order partial derivatives. z=ln(xy-sqrt(x^2-y^2)

so, do you mean

ln(xy-sqrt(x^2-y^2))
or
ln(xy)-sqrt(x^2-y^2)

I will assume the first, since the 2nd is much easier

∂z/∂x = 1/ln(xy-√(x^2-y^2)) * (y-x/√(x^2-y^2))

and similarly for y.

To find the first-order partial derivatives of z = ln(xy - sqrt(x^2 - y^2)), we need to find the partial derivative with respect to x and the partial derivative with respect to y.

Let's start by finding the partial derivative with respect to x (denoted as ∂z/∂x):

Step 1: Take the natural logarithm of both sides of the equation to simplify the expression:
ln(z) = ln(xy - sqrt(x^2 - y^2))

Step 2: Differentiate both sides of the equation with respect to x:
1/z * ∂z/∂x = y - 1/2 * (2x / sqrt(x^2 - y^2))
1/z * ∂z/∂x = y - x / sqrt(x^2 - y^2)

Step 3: Multiply both sides of the equation by z to isolate the partial derivative:
∂z/∂x = z * (y - x / sqrt(x^2 - y^2))

Now, let's find the partial derivative with respect to y (denoted as ∂z/∂y):

Step 1: Take the natural logarithm of both sides of the equation to simplify the expression:
ln(z) = ln(xy - sqrt(x^2 - y^2))

Step 2: Differentiate both sides of the equation with respect to y:
1/z * ∂z/∂y = x - 1/2 * (-2y / sqrt(x^2 - y^2))
1/z * ∂z/∂y = x + y / sqrt(x^2 - y^2)

Step 3: Multiply both sides of the equation by z to isolate the partial derivative:
∂z/∂y = z * (x + y / sqrt(x^2 - y^2))

So, the first-order partial derivatives are:
∂z/∂x = z * (y - x / sqrt(x^2 - y^2))
∂z/∂y = z * (x + y / sqrt(x^2 - y^2))

To find the first-order partial derivatives of the function \(z = \ln(xy - \sqrt{x^2 - y^2})\), we differentiate \(z\) with respect to each variable while treating the other variable as a constant.

Let's differentiate \(z\) with respect to \(x\) first:

1. Fixing \(y\) as a constant, differentiate each term individually:
- The derivative of \(\ln(u)\) with respect to \(x\) is \(\frac{1}{u} \cdot \frac{du}{dx}\).
- Applying this to the first term \(\ln(xy)\), we have \(\frac{1}{xy} \cdot \frac{d(xy)}{dx}\).
- Using the product rule for differentiation, \(\frac{d(xy)}{dx} = y + xy'\) where \(y\) is constant and \(y'\) is the derivative of \(x\) with respect to \(x\) (which is simply 1).
- Therefore, the derivative of the first term is \(\frac{1}{xy}(y + xy') = \frac{y}{xy} + \frac{x}{xy} = \frac{y}{x} + \frac{1}{y}\).

2. Now we need to differentiate the second term, \(-\ln(\sqrt{x^2 - y^2})\), with respect to \(x\).
- Applying the chain rule, the derivative of \(-\ln(\sqrt{x^2 - y^2})\) with respect to \(x\) is \(-\frac{1}{\sqrt{x^2 - y^2}} \cdot \frac{d(\sqrt{x^2 - y^2})}{dx}\).
- Differentiating \(\sqrt{x^2 - y^2}\) with respect to \(x\) using the chain rule, we get \(\frac{1}{2\sqrt{x^2 - y^2}} \cdot \frac{d(x^2 - y^2)}{dx}\).
- Using the power rule, the derivative of \(x^2 - y^2\) with respect to \(x\) is \(2x\).
- Therefore, the derivative of the second term is \(-\frac{1}{2\sqrt{x^2 - y^2}} \cdot \frac{2x}{\sqrt{x^2 - y^2}} = -\frac{x}{x^2 - y^2}\).

Thus, the first-order partial derivative of \(z\) with respect to \(x\) is \(\frac{y}{x} + \frac{1}{y} - \frac{x}{x^2 - y^2}\).

To find the first-order partial derivative of \(z\) with respect to \(y\), we follow the same steps while treating \(x\) as a constant:

1. Differentiating the first term \(\ln(xy)\) with respect to \(y\), we have \(\frac{1}{xy} \cdot \frac{d(xy)}{dy}\).
- Using the product rule, \(\frac{d(xy)}{dy} = x + x'y\) where \(x\) is constant and \(x'\) is the derivative of \(y\) with respect to \(y\) (which is simply 1).
- Therefore, the derivative of the first term is \(\frac{1}{xy}(x + x'y) = \frac{x}{xy} + \frac{y}{xy} = \frac{x}{y} + \frac{1}{x}\).

2. Differentiating the second term \(-\ln(\sqrt{x^2 - y^2})\) with respect to \(y\), we get 0 since it does not depend on \(y\).

Thus, the first-order partial derivative of \(z\) with respect to \(y\) is \(\frac{x}{y} + \frac{1}{x}\).