Partial differentiation equation help

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Homework Statement



derive (s2t3) / (rs2t3) with respect to s



The Attempt at a Solution



equation becomes s2t3*(rs2t3)-1

which becomes s2t3r-1s-2t-3

then just differantiate like a polynomial?

i tried this on an online partial differentiation calculator and it gave me an answer of:

(2s*t3(r*s2t3) - (s2t3)r*2s*t3)/(r*s2t3)2

where the heck do they get that from
 
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It simplifies more than that:
\frac{s^2^3}{rs^2t^3}= \frac{1}{r}= r^{-1} as long as s and t are not 0.

Its derivative with respect to s is 0.

(the rather strange formula you got is probably due to using the quotient rule. However, it should be easy to see that the numerator is of the for a- a= 0.)
 
thats odd because of the questions i have.

the entire problem is this.

f(r,s,t)=rln(rs2t3)

partially differentiate to find a) frss and b) frst

fr is what i gave in the previous post.. to be fr=(s2t3) / (rs2t3)


then if frs is 0 then both answers to the questions a) and b) are zero without even doing the 3rd differential.
that just seems odd?
 
But \partial_r [ r \ln (r s^2 t^3) ] = \ln ( r s^2 t^3 ) + 1, isn't it?
 
i thought:

derivitive of ln(u) = 1/u*u' so that derivitive became 1/(rs2t3)*(s2t3)
 
Of course, but you have a r in front of the logarithm multiplying it, so you need first to apply the product rule.
BTW: You can simplify your expression, 1/(r s² t³) *(s² t³) = 1/r. Which by multiplying with the r in front of the logarithm gives you the 1 in my result (second term of product rule).
 
ohhhh. i see. so product rule where r=u and ln(rs^t^3)=v.

helps if i derive right hey
 
:biggrin:
 

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