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- Partial derivatives
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Partial derivatives
Given the function $f(x,y)=x^2y^3-2xyz^3$ calculate the slope of the tangent line at the point $(1,5)$ in the direction of the axis $x$.
We have to calculate $$\dfrac{\delta f}{\delta x}$$ $$\dfrac{\delta f}{\delta x}=\dfrac{(1+y)(2x)-(x+y+xy)(2)}{(2x)^2}=$$ $$=\dfrac{2x+2xy-2x-2y-2xy}{2\cdot2\cdot x^2}=$$ $$=\dfrac{-y}{2x^2}$$
And now, the slope at $(1,5)$
$$\dfrac{\delta f(1,5)}{\delta x}=\dfrac{-5}{2\cdot1^2}=\dfrac{-5}{2}$$
The slope of the tangent line at the point $(1,5)$ in the direction of the axis $x$ is descendent, $\dfrac{-5}{2}$.
Given the function $f(x,y,z)=xy\cdot\ln(z)$ calculate the partial derivatives with respect to $x$, $y$ and $z$.
$$\dfrac{\delta f}{\delta x}=y\ln(z)$$
$$\dfrac{\delta f}{\delta y}=x\ln(z)$$
$$\dfrac{\delta f}{\delta z}=0\cdot\ln(z)+xy\cdot\dfrac{1}{z}=\dfrac{xy}{z}$$
$$\dfrac{\delta f}{\delta x}=y\ln(z)$$
$$\dfrac{\delta f}{\delta y}=x\ln(z)$$
$$\dfrac{\delta f}{\delta z}=0\cdot\ln(z)+xy\cdot\dfrac{1}{z}=\dfrac{xy}{z}$$