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	<title>Comments on: Calculus Partial Derivative?</title>
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		<title>By: nohglf</title>
		<link>http://fitnessone.net/blog/calculus-partial-derivative/comment-page-1/#comment-545</link>
		<dc:creator>nohglf</dc:creator>
		<pubDate>Tue, 06 Jan 2009 11:02:32 +0000</pubDate>
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		<description>For the first part you need to find

You need to find the partial derivative with respect to x,
the partial derivative with respect to y, and the partial derivative with respect to z.


With partial derivatives you treat all variables as constants unless it is the variable that the derivative is respect.

For example if you are trying to find the partial derivative with respect to x, all other variables such as y,z, or anything else is treated as a constant.


I have the following

partial-x = 2xy^3 + 2yz

partial-y = 3x^2y^2 + 2xz - 3z

 partial-z = 2xy -3y

You then after evalute each of these at the point (-2,1,2).

For question 2 the defintion of the total differential would
give

dw = (partial-x) dx + (partial-y) dy + (partial-z) dz

Where dw is the total differential.

Again you need to calculate partial derivatives.

Check: I have the following

(partial-x) = -e^ysin(x)

(partial-y) = e^y(cos x)

(partial-z) = 2z

Note: The problem you gave does not have values for dx, dy, 
and dz so you would leave those terms in your equation.

Question 3:

You can approach 2 different ways. You transform the original equation expressed in terms of t and take dw/dt directly (usually not an efficient approach). Or you could use the chain rule. The chain rule would involve this equation.

dw/dt = (dx/dt)(partial-w / partial x) + (dy/dt) (partial - w/ partial-y) + (dz/dt) (partial-w/ partial-z)

You can actually check to see if you did the chain rule correctly by plugging the expressions for &#039;t&#039; back into the original equation and calculating dw/dt directly (you should get the same answer).</description>
		<content:encoded><![CDATA[<p>For the first part you need to find</p>
<p>You need to find the partial derivative with respect to x,<br />
the partial derivative with respect to y, and the partial derivative with respect to z.</p>
<p>With partial derivatives you treat all variables as constants unless it is the variable that the derivative is respect.</p>
<p>For example if you are trying to find the partial derivative with respect to x, all other variables such as y,z, or anything else is treated as a constant.</p>
<p>I have the following</p>
<p>partial-x = 2xy^3 + 2yz</p>
<p>partial-y = 3x^2y^2 + 2xz &#8211; 3z</p>
<p> partial-z = 2xy -3y</p>
<p>You then after evalute each of these at the point (-2,1,2).</p>
<p>For question 2 the defintion of the total differential would<br />
give</p>
<p>dw = (partial-x) dx + (partial-y) dy + (partial-z) dz</p>
<p>Where dw is the total differential.</p>
<p>Again you need to calculate partial derivatives.</p>
<p>Check: I have the following</p>
<p>(partial-x) = -e^ysin(x)</p>
<p>(partial-y) = e^y(cos x)</p>
<p>(partial-z) = 2z</p>
<p>Note: The problem you gave does not have values for dx, dy,<br />
and dz so you would leave those terms in your equation.</p>
<p>Question 3:</p>
<p>You can approach 2 different ways. You transform the original equation expressed in terms of t and take dw/dt directly (usually not an efficient approach). Or you could use the chain rule. The chain rule would involve this equation.</p>
<p>dw/dt = (dx/dt)(partial-w / partial x) + (dy/dt) (partial &#8211; w/ partial-y) + (dz/dt) (partial-w/ partial-z)</p>
<p>You can actually check to see if you did the chain rule correctly by plugging the expressions for &#8216;t&#8217; back into the original equation and calculating dw/dt directly (you should get the same answer).</p>
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