# If w=sqrt(x^2+y^2) x=ln(rs+tu) y=t/u(cos(rs))... find dw/dt, dw/ds, dw/dr, dw/du.I'm confused on what the difference between dw/dt & dw/ds is and what exactly is dw? Thanks!

rcmath | Certified Educator

We are going to assume that t,s,r, and u are independant from each other.

For sake of simplicity, I am going to use the ' symbol first, then differentiate appropriately.

`w=sqrt(x^2+y^2) =>`

`w'=1/2*(x^2+y^2)^(-1/2)*(x^2+y^2)'`

`w'=-((x^2+y^2)')/(2*sqrt(x^2+y^2))`

`w'=-(2x*x'+2y*y')/(2sqrt(x^2+y^2))`

`w'=-(x*x'+y*y')/sqrt(x^2+y^2)`

If we differentiate with respect to t, the other 3 variables are going to be treated like constants, so

`(dx)/(dt)=(-u)/(rs+tu)`

`(dy)/(dt)=1/(ucos(rs))`

Hence`(dw)/(dt)=`

`-[ln(rs+tu)*(-u/(rs+tu))+(t/ucos(rs))*(1/ucos(rs))]/sqrt(ln^2(rs+tu)+t^2/(ucos(rs))^2)`

`=[u^2ln(rs+tu)+(rs+tu)*t*cos^2(rs)]/[u*(rs+tu)*sqrt(ln^2(rs+tu)+t^2/(ucos(rs))^2))`

To differentiate with respect to u, you treat t, r, and s like constant and you differentiate appropriately.