find $\frac{dy}{dx}$ by implicit differentiation. $sin(x)+cos(y)=3x - 8y$ $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $\sin(x)+\cos(y)$ and $3x - 8y$ with respect to $x$. $\frac{d}{dx}(\sin(x)+\cos(y))=\frac{d}{dx}(3x - 8y)$ Using the sum - rule of differentiation $\frac{d}{dx}(u + v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)$, we get $\frac{d}{dx}(\sin(x))+\frac{d}{dx}(\cos(y))=\frac{d}{dx}(3x)-\frac{d}{dx}(8y)$. We know that $\frac{d}{dx}(\sin(x))=\cos(x)$, $\frac{d}{dx}(3x)=3$ and by the chain - rule $\frac{d}{dx}(\cos(y))=-\sin(y)\frac{dy}{dx}$, $\frac{d}{dx}(8y)=8\frac{dy}{dx}$. So, $\cos(x)-\sin(y)\frac{dy}{dx}=3 - 8\frac{dy}{dx}$.

Step2: Isolate $\frac{dy}{dx}$

Move all terms with $\frac{dy}{dx}$ to one side: $8\frac{dy}{dx}-\sin(y)\frac{dy}{dx}=3-\cos(x)$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(8 - \sin(y))=3-\cos(x)$. Then $\frac{dy}{dx}=\frac{3-\cos(x)}{8 - \sin(y)}$.

Answer:

$\frac{3-\cos(x)}{8 - \sin(y)}$