Get the answer to "Find both first partial derivatives. z=ln⁡(x+y)x−y"

permaneceerc

Answered question

2021-02-17

Find both first partial derivatives.

z = \frac{\ln (x + y)}{x - y}

Margot Mill

Skilled2021-02-19Added 106 answers

Step 1
Given Data:
Function:

z = \ln \frac{(x + y)}{(x - y)}

Rewrite the given function.

z = \ln (x + y) - \ln (x - y)

The partial derivative of the above function with respect to x is,

\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (\ln (x + y) - \ln (x - y))

= \frac{1}{x + y} - \frac{1}{x - y}

= \frac{x - y - (x + y)}{x + y (x - y)}

= \frac{x - y - x - y}{x^{2} - y^{2}}

= \frac{- 2 y}{x^{2} - y^{2}}

Step 2
The partial derivative of the function with respect to y is,

\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (\ln (x + y) - \ln (x - y))

= \frac{1}{x + y} - \frac{(- 1)}{x - y}

= \frac{x - y + (x + y)}{x + y (x - y)}

= \frac{x - y + x + y}{x^{2} - y^{2}}

= \frac{2 x}{x^{2} - y^{2}}

Thus, the partial derivative with respect to x is

\frac{- 2 y}{x^{2} - y^{2}}

and the partial derivative with respect y is

\frac{2 x}{x^{2} - y^{2}}

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