Partial Derivatives

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Objectives:

1.: Understand how to compute a partial derivative.
2.: Know what the partial derivative represents geometrically.

Recap Video

The following video recaps the main ideas of the section.

Problems

If $f(x,y) = y^2+6xy$ , then: $f_x = \answer {6y},\quad f_y = \answer {2y+6x}.$

If $f(x,y) = \sin (xy)+yz+\frac {x}{z}$ , then: $f_x = \answer {y\cos (xy)+\frac {1}{z}},\quad f_y = \answer {x\cos (xy)+z},\quad f_z = \answer {y-\frac {x}{z^2}}.$

If $f(x,y) = xe^{xy}$ , then: $f_x = \answer {xye^{xy}+e^{xy}},\quad f_y = \answer {x^2e^{xy}}.$

If $f(x,y) = \ln (xy)$ , then $\frac {\partial ^2 f}{\partial y \partial x} = \answer {0}.$

If $f(x,y) = y^2x^3+x^2e^y$ , then:

$f_{xx} = \answer {6xy^2+2e^y},$
$f_{xy} = \answer {6x^2y+2xe^y},$
$f_{yx} = \answer {6x^2y+2xe^y},$
$f_{yy} = \answer {2x^3+x^2e^y}.$

If $f(x,y) = xe^{\sin (yz)}$ , then $f_{yzxyzx} = \answer {0}$ .

Use Clairaut’s theorem to reorder the partials.

The contour map of function $z = f(x,y)$ is shown below.

The partial $f_x(P)$ is positivenegativezero and $f_y(P)$ is positivenegativezero .

What are the $z$ -values doing as you move in the $x$ and $y$ directions?