How to differentiate x^(cosx) = y^(sinx) with respect to x


by styxrihocc
Tags: differentiate, respect, xcosx, ysinx
styxrihocc
styxrihocc is offline
#1
Apr18-12, 12:34 PM
P: 10
1. The problem statement, all variables and given/known data

Differentiate x^(cosx) = y ^(sinx) with respect to x

2. Relevant equations



3. The attempt at a solution
I tried using natural logs but im not sure if its correct, if it's wrong please point me to the right direction, thanks

x^(cosx) = y^(sinx)
ln x^(cosx) = ln y ^(sinx)
ln x (cosx) = ln y (sinx)
cosx/x - sinx lnx = cosx lny +sinx/y (dy/dx)
cosx/x - sinx lnx - cosx lny = sinx/y (dy/dx)
(cosx/x - sinx lnx - cosx lny) / (sinx/y) = dy/dx
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
LCKurtz
LCKurtz is offline
#2
Apr18-12, 01:06 PM
HW Helper
Thanks
PF Gold
LCKurtz's Avatar
P: 7,187
Quote Quote by styxrihocc View Post
1. The problem statement, all variables and given/known data

Differentiate x^(cosx) = y ^(sinx) with respect to x

2. Relevant equations



3. The attempt at a solution
I tried using natural logs but im not sure if its correct, if it's wrong please point me to the right direction, thanks

x^(cosx) = y^(sinx)
ln x^(cosx) = ln y ^(sinx)
ln x (cosx) = ln y (sinx)
cosx/x - sinx lnx = cosx lny +sinx/y (dy/dx)
cosx/x - sinx lnx - cosx lny = sinx/y (dy/dx)
(cosx/x - sinx lnx - cosx lny) / (sinx/y) = dy/dx
Your work is correct. You could be slightly less ambiguous with parentheses though.
styxrihocc
styxrihocc is offline
#3
Apr18-12, 01:17 PM
P: 10
Thanks.


Register to reply

Related Discussions
integral 1/(cosx -sinx) Calculus & Beyond Homework 7
sinx+cosx> = 1 Calculus 20
lim (sinx-cosx)/(pi-4x) where x-->pi/4 Calculus & Beyond Homework 9
solve cosx^4-sinx^4 Precalculus Mathematics Homework 2
Differentiate y=x sinx cosx Differential Equations 6