Partial Derivatives

Topics:

• Be able to compute first-order and second-order partial derivatives.
• Be able to perform implicit partial differentiation.
• Be able to solve various word problems involving rates of change, which use partial derivatives.

For a function $z=f(x,y)$ we define:

The partial derivative $f$ with respect to $x$ at the point $P(a,b)$ as

$$\underset{\text{Holding ? constant}}{\underbrace{f_x(a,b)=\underset{h\to 0}{\lim }\frac{f(a+h,b)-f(a,b)}{h}}}$$

And with respect to $y$ at the same

$$f_y(a,b)=\underset{h\to 0}{\lim }\frac{f(a,b+h)-f(a,b)}{h}$$

Notation:

$$\begin{array}{rl}{f}_x(x,y)& =\frac{\partial f}{\partial x}=\frac{\partial z}{\partial x}=\dots \\ f_y(x,y)& =\frac{\partial f}{\partial y}=\frac{\partial z}{\partial y}=\dots \end{array}$$

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Example

$$\begin{array}{rl}f(x,y)=2x^2+5xy+4y^6& \\ & \\ f_x(x,y)& =\frac{\partial f}{\partial x}=4x+5y+0\text{ (hold y constant)}\\ & \\ f_y(x,y)& =\frac{\partial f}{\partial y}=0+5x+24y^5\text{ (hold x constant)}\end{array}$$

Another one

$$\begin{array}{rl}& \\ h(x,y)=(2z-4xy{)}^5& \\ & \\ h_x(x,y)& =5(2x-4xy{)}^4\cdot (2-4y)\\ & \\ h_y(x,y)& =5(2x-4xy{)}^4\cdot (0-4x)\end{array}$$

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Once you find a partial derivative, you can differentiate that partial derivative function again with respect to one of the variables. $f_{xx}=\frac{{\partial }^{2}f}{\partial x^2}$ or $f_{xy}=\frac{{\partial }^{2}f}{\partial y\partial x}$ (notation is WEIRD and BACKWARDS)

Mixed partials (in most cases) turn out to be the same thing (for us, where we are, what we use this for, in our level). This is called Clairaut’s Theorem.

Clairaut’s Theorem:

Suppose $f$ is defined on a disk $D$ that contains the point $P(a,b)$. If $f_{xy}$ and $f_{yx}$ are both continuous on $D$, then $f_{xy}(a,b)=f_{yx}(a,b)$.

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