What is the derivative of #ln y=e^y sinx#?
1 Answer
Notice how these functions are functions of
#(df(y))/(dx) = (df(y))/(dy)*(dy)/(dx)#
Therefore, you can differentiate as normal, keeping the variable as
#d/(dx)[lny] = d/(dx)[e^y sinx]#
#1/y (dy)/(dx) = (e^y*cosx) + (sinx*e^y (dy)/(dx))#
(Product Rule on the right side)
Now we should isolate
#1/y (dy)/(dx) - e^ysinx (dy)/(dx) = e^ycosx#
Factor:
#[1/y - e^ysinx] (dy)/(dx) = e^ycosx#
Divide:
#(dy)/(dx) = (e^ycosx)/[1/y - e^ysinx]#
Getting rid of the fractions within fractions by multiplying by
#= color(blue)((e^y ycosx)/[1 - e^y ysinx])#