$\displaystyle (xy+1)^3 = x - y^2 + 8$Find the derivative of $\displaystyle (xy+1)^3 = x - y^2 + 8$ using implicit differentiation. Do not simplify.

$\displaystyle 3(xy+1)^2 \cdot (y + x \cdot \frac{dy}{dx}) = 1 - 2y \cdot \frac{dy}{dx} $

$\displaystyle y + x \cdot \frac{dy}{dx} + \frac{2y \cdot \frac{dy}{dx}}{3(xy+1)^2} = \frac {1}{3(xy+1)^2}$

$\displaystyle x \cdot \frac{dy}{dx} + \frac{2y \cdot \frac{dy}{dx}}{3(xy+1)^2} = \frac {1}{3(xy+1)^2} -y$

$\displaystyle \frac{dy}{dx}[x + \frac{2y}{3(xy+1)^2}] = \frac {1}{3(xy+1)^2} -y$

$\displaystyle \frac{dy}{dx} = \frac {\frac {1}{3(xy+1)^2} -y}{x + \frac{2y}{3(xy+1)^2}}$

The question stated not to simplify, so this would be my final answer.

$\displaystyle \sin(xy) + x = e^y$Find the derivative of $\displaystyle \sin(xy) + x = e^y$ using implicit differentiation. Do not simplify.

$\displaystyle \cos(xy) \cdot (y + x \cdot \frac{dy}{dx}) + 1 = e^y \cdot \frac{dy}{dx}$

$\displaystyle \cos(xy) \cdot (y + x \cdot \frac{dy}{dx}) - e^y \cdot \frac{dy}{dx} = -1$

$\displaystyle x \cdot \frac{dy}{dx} - \frac{e^y \cdot \frac{dy}{dx}}{\cos(xy)} = \frac{-1}{\cos(xy)} - y$

$\displaystyle \frac{dy}{dx}[x - \frac{e^y}{\cos(xy)}] = \frac{-1}{\cos(xy)} - y$

$\displaystyle \frac{dy}{dx} = \frac{\frac{-1}{\cos(xy)} - y}{[x - \frac{e^y}{\cos(xy)}]}$

I opted to not use logarithmic differentiation for some reason. Is this solution still valid?

Thanks a lot!