Problems from Partial derivatives

Development:

We have to calculate $${\displaystyle \frac{\delta f}{\delta x}}$$ $${\displaystyle \frac{\delta f}{\delta x}}={\displaystyle \frac{(1+y)(2x)-(x+y+xy)(2)}{(2x{)}^2}}=$$ $$={\displaystyle \frac{2x+2xy-2x-2y-2xy}{2\cdot 2\cdot x^2}}=$$ $$={\displaystyle \frac{-y}{2x^2}}$$

And now, the slope at $(1,5)$

$${\displaystyle \frac{\delta f(1,5)}{\delta x}}={\displaystyle \frac{-5}{2\cdot 1^2}}={\displaystyle \frac{-5}{2}}$$

Solution:

The slope of the tangent line at the point $(1,5)$ in the direction of the axis $x$ is descendent, ${\displaystyle \frac{-5}{2}}$.

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Development:

$${\displaystyle \frac{\delta f}{\delta x}}=y\ln (z)$$

$${\displaystyle \frac{\delta f}{\delta y}}=x\ln (z)$$

$${\displaystyle \frac{\delta f}{\delta z}}=0\cdot \ln (z)+xy\cdot {\displaystyle \frac{1}{z}}={\displaystyle \frac{xy}{z}}$$

Solution:

${\displaystyle \frac{\delta f}{\delta x}}=y\ln (z)$

${\displaystyle \frac{\delta f}{\delta y}}=x\ln (z)$

${\displaystyle \frac{\delta f}{\delta z}}=0\cdot \ln (z)+xy\cdot {\displaystyle \frac{1}{z}}={\displaystyle \frac{xy}{z}}$

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