How do you determine dy/dx if y=x+tan(xy)y=x+tan(xy)?

1 Answer
May 23, 2018

dy/dx=frac{cos^2(xy)+y}{cos^2(xy)-x}dydx=cos2(xy)+ycos2(xy)x

Explanation:

we get by the chain rule:
y'=1+1/cos^2(xy)(y+xy')

dy/dx = 1 + y/cos^2(xy) + dy/dx (x/cos^2(xy))

dy/dx - dy/dx(x/cos^2(xy)) = 1+ysec^2(xy)

dy/dx * (1-frac{x}{cos^2(xy)}) = 1+ysec^2(xy)

dy/dx = frac{1+ysec^2(xy)}{1 - xsec^2(xy)}

Multiplying by cos^2(xy) over cos^2(xy) we get

dy/dx=frac{cos^2(xy)+y}{cos^2(xy)-x}