Find dy/dx.
e^(xy)=sin(x)
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
To find dy/dx, we need to differentiate both sides of the equation with respect to x using the chain rule.
The given equation is e^(xy) = sin(x).
Differentiating both sides with respect to x:
d/dx(e^(xy)) = d/dx(sin(x)).
Using the chain rule, the derivative of e^(xy) with respect to x is:
d/dx(e^(xy)) = d/d(xy) * (xy) * d/dx(x).
= y * e^(xy).
The derivative of sin(x) with respect to x is:
d/dx(sin(x)) = cos(x).
So, y * e^(xy) = cos(x).
To solve for dy/dx, we divide both sides by y * e^(xy):
(dy/dx) = cos(x) / (y * e^(xy)).
Therefore, the derivative dy/dx is equal to cos(x) divided by (y * e^(xy)).