Partial differentiation equation help

[ProPatto16]

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

 

[HallsofIvy]

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.)

 

[ProPatto16]

thats odd because of the questions i have.

the entire problem is this.

f(r,s,t)=rln(rs²t³)

partially differentiate to find a) f_rss and b) f_rst

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


then if f_rs is 0 then both answers to the questions a) and b) are zero without even doing the 3rd differential.
that just seems odd?

 

[grey_earl]

But $$\partial_r [ r \ln (r s^2 t^3) ] = \ln ( r s^2 t^3 ) + 1$$, isn't it?

 

[ProPatto16]

i thought:

derivitive of ln(u) = 1/u*u' so that derivitive became 1/(rs²t³)*(s²t³)

 

[grey_earl]

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).

 

[ProPatto16]

ohhhh. i see. so product rule where r=u and ln(rs^t^3)=v.

helps if i derive right hey

 

[grey_earl]