Question

use implicit differentiation to find $\frac{dy}{dx}$. $cos(3y^{2}) - 6x = 2e^{y}$ $\frac{dy}{dx}=square$ (use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Differentiate both sides

Differentiate $\cos(3y^{2})-6x = 2e^{y}$ with respect to $x$. The derivative of $\cos(3y^{2})$ using the chain - rule is $-\sin(3y^{2})\cdot6y\frac{dy}{dx}$, the derivative of $-6x$ is $-6$, and the derivative of $2e^{y}$ is $2e^{y}\frac{dy}{dx}$. So we have:
$-\sin(3y^{2})\cdot6y\frac{dy}{dx}-6 = 2e^{y}\frac{dy}{dx}$

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

Move all terms with $\frac{dy}{dx}$ to one side:
$-\sin(3y^{2})\cdot6y\frac{dy}{dx}-2e^{y}\frac{dy}{dx}=6$
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(- 6y\sin(3y^{2})-2e^{y}) = 6$
Then $\frac{dy}{dx}=\frac{6}{-6y\sin(3y^{2}) - 2e^{y}}=-\frac{3}{3y\sin(3y^{2})+e^{y}}$

Answer:

$-\frac{3}{3y\sin(3y^{2})+e^{y}}$