Partial Derivates

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EllieJones

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Hello, I missed my last lesson about partial derivates, so I'm kinda clueless about this subject and I'm stuck with this problem. I tried solving it, but I'm sure my answer is 100% wrong. Can anyone tell me what I'm doing wrong?
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Here's a better way to think of partial derivatives.
Remember the quadratic ax2+bx+c? You can take the partial derivative with respect to x, and get 2ax+b.

For this problem, you're incorrectly taking the derivative. Recall that for the product rule,
[MATH](uv)'=uv'+u'v[/MATH]You've done
[MATH](uv)'=u'v'[/MATH] which is incorrect.
 
EllieJones said:
View attachment 23308
Hello, I missed my last lesson about partial derivates, so I'm kinda clueless about this subject and I'm stuck with this problem. I tried solving it, but I'm sure my answer is 100% wrong. Can anyone tell me what I'm doing wrong?
View attachment 23309P
Click to expand...
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It will be simpler for you, if you simplify the function by multiplying it out and then differentiate.

f(r,s,t) = r4*s*t + r*s3*t + r*s*t5

df/dr = 4*s*t*r3 + s3*t + s*t5

df/ds = r4*t + r*(3 * s2)*t + r*t5

......... continue......
 
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Your mistake was that you did not use the product rule as implied by "anon_hedgepig" (I love that user name!). \(\displaystyle rst(r^3+ s^2+ t^4)]_s= (rst)_s(r^3+ s^2+ t^4)+ (rst)(r^3+ s^2+ t^4)_s= (rt)(r^3+ s^2+ t^4)+ (rst)(2s)= r^4t+rs^2t+rt^5+2rs^2t=r^4t+ 3rs^2t+ rt^5\).

That is the same as if you wrote the function as \(\displaystyle r^4st+ rs^3t+ rst^5\) so the derivative is \(\displaystyle r^4t+ 3rs^2t+ rt^5\).
 
You understand perfectly how to do partial derivatives (treat r and t as constants). What you do not know how to do is to compute the derivative of a product which is NOT the product of the derivative.
 
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