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 $$x$$.
See development and solution
Development:
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 $$(1,5)$$
$$$\dfrac{\delta f(1,5)}{\delta x}=\dfrac{-5}{2\cdot1^2}=\dfrac{-5}{2}$$$
Solution:
The slope of the tangent line at the point $$(1,5)$$ in the direction of the axis $$x$$ is descendent, $$\dfrac{-5}{2}$$.