Question

f(x,y) = cos[ln(x/y)]
What is this differentiated respect to y? Please show all working.

Answer 1

use chain rule and log property

ln(x/y) = ln[(y/x)^-1] = -ln(y/x)
u = -ln(y/x)
du = -1/y

d/dy cos(u) = -sin(u)*du = sin[ln(x/y)]/y

Answer 2

Differentiating the function directly going from the outermost operator which is cos we get
Wait wait i will type the answer first and explain
\[\sin (\ln (x/y))*(y/x)*(-x/y^2)\]
Here first thing is the differential of cos() and the second term is the differential of ln() from analogy that differential of\[(\ln (x))\prime=1/x\]
And the third term is the inner derivative of (x/y) Since x is treated as a constant.
And finally the answer is\[\sin (\ln (x/y))*(y/x)*(-x/y^2)=-\sin(\ln(x/y))/y\]