Question

Obtain all the first and second order partial derivatives of the function: f ( x, y ) =ln ( 1 + xy ^ 2)

Answer 1

Answer on Question#50737

1) Obtain all the first and second order partial derivatives of the function: $f(x, y) = \ln(1 + xy^2)$

Solution. Let's compute first order partial derivatives of the origin function:

$$\frac {\partial}{\partial x} f (x, y) = \frac {y ^ {2}}{1 + x y ^ {2}}; \quad \frac {\partial}{\partial y} f (x, y) = \frac {2 x y}{1 + x y ^ {2}}$$

Let's compute second order partial derivatives of the origin function:

$$\frac {\partial^ {2}}{\partial x ^ {2}} f (x, y) = \frac {\partial}{\partial x} \left(\frac {y ^ {2}}{1 + x y ^ {2}}\right) = \frac {\partial}{\partial x} \left(y ^ {2} (1 + x y ^ {2}) ^ {- 1}\right) = - y ^ {2} (1 + x y ^ {2}) ^ {- 2} y ^ {2} = \frac {- y ^ {4}}{\left(1 + x y ^ {2}\right) ^ {2}};$$ $$\frac {\partial^ {2}}{\partial x \partial y} f (x, y) = \frac {\partial^ {2}}{\partial y \partial x} f (x, y) = \frac {\partial}{\partial y} \left(\frac {y ^ {2}}{1 + x y ^ {2}}\right) = \frac {2 y (1 + x y ^ {2}) - 2 x y ^ {3}}{(1 + x y ^ {2}) ^ {2}} = \frac {2 y}{(1 + x y ^ {2}) ^ {2}};$$ $$\frac {\partial^ {2}}{\partial y ^ {2}} f (x, y) = \frac {\partial}{\partial y} \left(\frac {2 x y}{1 + x y ^ {2}}\right) = \frac {2 x (1 + x y ^ {2}) - 4 x ^ {2} y ^ {2}}{(1 + x y ^ {2}) ^ {2}} = \frac {2 x - 2 x ^ {2} y ^ {2}}{(1 + x y ^ {2}) ^ {2}}.$$

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