How do you implicitly differentiate #-1=xy+e^ysec(x/y) #?
1 Answer
I got
#(dy)/(dx) = (-y - 1/ye^ysec(x/y)tan(x/y))/[x - e^ysec(x/y)(x/y^2tan(x/y) + 1)]#
Since
#d/(dx)[-1] = d/(dx)[xy + e^ysec(x/y)]#
#0 = (x(dy)/(dx) + y) + (e^yd/(dx)[sec(x/y)] + sec(x/y)d/(dx)[e^y])#
#= (x(dy)/(dx) + y) + (e^ysec(x/y)tan(x/y)*d/(dx)[x/y] + sec(x/y)e^y(dy)/(dx))#
#= x(dy)/(dx) + y + [e^ysec(x/y)tan(x/y)*(-x/y^2(dy)/(dx)+1/y) + sec(x/y)e^y(dy)/(dx)]#
Now we simplify and isolate
#0 = x(dy)/(dx) + y + [-x/y^2e^ysec(x/y)tan(x/y)(dy)/(dx) + 1/ye^ysec(x/y)tan(x/y) + sec(x/y)e^y(dy)/(dx)]#
#0 = x(dy)/(dx) + y - x/y^2e^ysec(x/y)tan(x/y)(dy)/(dx) + 1/ye^ysec(x/y)tan(x/y) + sec(x/y)e^y(dy)/(dx)#
It's all multiplied out, so separate your variables.
#- y - 1/ye^ysec(x/y)tan(x/y) = x(dy)/(dx) - x/y^2e^ysec(x/y)tan(x/y)(dy)/(dx) + sec(x/y)e^y(dy)/(dx)#
#- y - 1/ye^ysec(x/y)tan(x/y) = [x - x/y^2e^ysec(x/y)tan(x/y) + sec(x/y)e^y](dy)/(dx)#
#(dy)/(dx) = (-y - 1/ye^ysec(x/y)tan(x/y))/[x - x/y^2e^ysec(x/y)tan(x/y) + sec(x/y)e^y]#
Simplifying this further, we'd get:
#color(blue)((dy)/(dx) = (-y - 1/ye^ysec(x/y)tan(x/y))/[x - e^ysec(x/y)(x/y^2tan(x/y) + 1)])#