Find dy/dx.
e^(xy)=sin(x)
2 answers
Sam/Maggie/Kate -- please use the same name for your posts.
better use implicit differentiation:
e^(xy) = sinx
e^(xy)(y + xy') = cosx
x e^(xy) y' = cosx - y e^(xy)
y' = (cosx - y e^(xy))/(x e^(xy))
but e^(xy) = sinx, so
y' = (cosx - y*sinx)/(x*sinx)
y' = (cotx - y)/x
or, using logs,
e^(xy) = sinx
xy = ln(sinx)
y + xy' = 1/sinx * cosx
xy' = cotx - y
y' = (cotx - y)/x
e^(xy) = sinx
e^(xy)(y + xy') = cosx
x e^(xy) y' = cosx - y e^(xy)
y' = (cosx - y e^(xy))/(x e^(xy))
but e^(xy) = sinx, so
y' = (cosx - y*sinx)/(x*sinx)
y' = (cotx - y)/x
or, using logs,
e^(xy) = sinx
xy = ln(sinx)
y + xy' = 1/sinx * cosx
xy' = cotx - y
y' = (cotx - y)/x
1 answer