14.3 - Partial Derivatives

Abstract: We know that derivatives on the 2d plane find the slope of the line at that point, but what does the derivative find on the 3d plane?

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Finding the Slope on the 3D Plane

Definition: X axis

Partial Derivative with respect to x means the slope of the surface in the direction of the positive x-axis

Definition: Y axis

Partial derivative with respect to Y means the slope in the direction of positive y-axis

We can also find the slope in the direction of the z-axis by taking the partial derivative with respect to z.

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Find the Partial Derivative with respect to X and Y

Definition: Partial Derivative with respect to X

The partial derivative with respect to x is the slope of the surface in the direction of the positive x-axis

Definition: Partial Derivative with respect to Y

The partial derivative with respect to y is the slope of the surface in the direction of the positive y-axis

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Example: Find $f_x$ and $f_y$

1.

We are given:

$f(x,y)=x^2{y}^5$

Finding $f$ with respect to x:

$f_x=y^5(2x)=2x{y}^5$

Finding $f$ with respect to y:

$f_y=5y^4{x}^2$

2.

Finding $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$

$z=\frac{y^2}{(1+x^2{)}^3}$

Lets solve.

$\frac{\partial z}{\partial x}=y^2\frac{\partial }{\partial x}\frac{1}{(1+x^2)}=y^2\frac{\partial }{\partial x}(1+x{)}^{-3}$

This then becomes:

$=y^2(-3)(1+x^2{)}^{-4}(2x)=-\frac{6x{y}^2}{(1+x^2{)}^4}$

Solving for $\frac{\partial z}{\partial y}$

$\frac{\partial z}{\partial y}=\frac{2y}{(1+x^2{)}^3}$

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Practice Problems

Given:

$$f={\int }_x^y\cos (e^t)dt=-{\int }_y^x\cos (e^t)dt$$ $$f_y=\frac{\partial }{\partial y}\left[{\int }_x^y\cos (e^t)dt\right]$$ $$(f_x{)}_x=f_{xx}$$ $$f=x^2+y^2+e^x$$ $$f_{xx}=6x+y^2{e}^x$$ $$f_{xy}=(f_x{)}_y=2y{e}^x$$