Determine the partial derivative with respect to $x$ and with respect to $y$ of the function
$$z(x,y) = \ln(x^2 + 2y^2 + 13).$$
For the determination of both partial derivatives it is useful to write $z(x,y)$ as
$$z(x,y) = u\big(v(x,y)\big) \qquad \text{with} \qquad u(v) = \ln(v) \qquad \text{and} \qquad v(x,y) = x^2 + 2y^2 + 13.$$
For both partial derivatives we use the chain rule.
For the determination of the partial derivative with respect to $x$ we need $v'_x(x,y)$. We treat $y$ and therefore also $2y^2$ as a constant; we obtain:
$$v'_x(x,y) = 2x + 0 + 0 = 2x.$$
Then the partial derivative of $z(x,y)$ with respect to $x$ is:
$$z'_x(x,y) = u'(v(x,y)) \cdot v'_x(x,y) = \dfrac{1}{v(x,y)} \cdot 2x = \dfrac{2x}{x^2 + 2y^2 + 13}.$$
For the determination of the partial derivative with respect to $y$ we need $v'_y(x,y)$. We treat $x$ and therefore also $x^2$ as a constant; we obtain:
$$v'_x(x,y) = 0 + 2\cdot 2y + 0 = 4y.$$
Then the partial derivative of $z(x,y)$ with respect to $y$ is:
$$z'_y(x,y) = u'(v(x,y)) \cdot v'_y(x,y) = \dfrac{1}{v(x,y)} \cdot 4y = \dfrac{4y}{x^2 + 2y^2 + 13}.$$
$$z(x,y) = \ln(x^2 + 2y^2 + 13).$$
For the determination of both partial derivatives it is useful to write $z(x,y)$ as
$$z(x,y) = u\big(v(x,y)\big) \qquad \text{with} \qquad u(v) = \ln(v) \qquad \text{and} \qquad v(x,y) = x^2 + 2y^2 + 13.$$
For both partial derivatives we use the chain rule.
For the determination of the partial derivative with respect to $x$ we need $v'_x(x,y)$. We treat $y$ and therefore also $2y^2$ as a constant; we obtain:
$$v'_x(x,y) = 2x + 0 + 0 = 2x.$$
Then the partial derivative of $z(x,y)$ with respect to $x$ is:
$$z'_x(x,y) = u'(v(x,y)) \cdot v'_x(x,y) = \dfrac{1}{v(x,y)} \cdot 2x = \dfrac{2x}{x^2 + 2y^2 + 13}.$$
For the determination of the partial derivative with respect to $y$ we need $v'_y(x,y)$. We treat $x$ and therefore also $x^2$ as a constant; we obtain:
$$v'_x(x,y) = 0 + 2\cdot 2y + 0 = 4y.$$
Then the partial derivative of $z(x,y)$ with respect to $y$ is:
$$z'_y(x,y) = u'(v(x,y)) \cdot v'_y(x,y) = \dfrac{1}{v(x,y)} \cdot 4y = \dfrac{4y}{x^2 + 2y^2 + 13}.$$