# Solve for $\frac{dy}{dx}$ of a trigonometric function after implicit differentiation

I'm supposed to implicitly differentiate $\sin(x+y)=2x-2y$. I've already taken the first derivative and got

$$\left(\frac{dy}{dx}+1\right)\cdot\cos(y+x)=-2\left(\frac{dy}{dx}-1\right)$$

www.derivative-calculator.net says solving this equation for $\frac{dy}{dx}$ equals

$$\frac{dy}{dx}=-\frac{\cos(y+x)-2}{\cos(y+x)+2}$$

But I'm lost when it comes to the algebra used to rewrite the equation in terms of $\frac{dy}{dx}$. Any help getting from point A to B will be greatly appreciated.

• These 2 equations are trivially equal. Mar 8, 2016 at 14:50

$$\left(\frac{dy}{dx}+1\right)\cdot\cos(y+x)=-2\left(\frac{dy}{dx}-1\right)$$ $$\frac{dy}{dx}\cos(y+x)+\cos(y+x)=-2\frac{dy}{dx}+2$$ $$\frac{dy}{dx}\cos(y+x)+2\frac{dy}{dx}=2-\cos(y+x)$$ $$\frac{dy}{dx}(\cos(y+x)+2)=-(-2+\cos(y+x))$$ $$\frac{dy}{dx}=-\frac{\cos(y+x)-2}{\cos(y+x)+2}$$

So, $$\sin(x+y) = 2(x-y)$$ thus $$\cos(x+y)+\frac{dy}{dx}\cos(x+y)=2-2\frac{dy}{dx}$$ Or $$\cos(x+y)\left(1+\frac{dy}{dx}\right)=2\left(1-\frac{dy}{dx}\right)$$ Re-arranging $$\cos(x+y)\frac{dy}{dx}+2\frac{dy}{dx}=2-\cos(x+y)$$ Hence, $$\frac{dy}{dx}(2+\cos(x+y) = -(\cos(x+y)-2)$$ From which $$\frac{dy}{dx} = -\frac{\cos(x+y)-2}{\cos(x+y)+2}$$