y sin 3x , y³ cos x+5=0 find dy/dx

To find $\frac{dy}{dx}$ for the equation $y^3 \cos x + 5 = 0$, we will use implicit differentiation. Step 1: Differentiate both sides of the equation with respect to $x$. $$ \frac{d}{dx}(y^3 \cos x + 5) = \frac{d}{dx}(0) $$ Step 2: Apply the product rule to $y^3 \cos x$. The product rule states $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = y^3$ and $v = \cos x$. Then $u' = \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$ (using the chain rule). And $v' = \frac{d}{dx}(\cos x) = -\sin x$. The derivative of the constant $5$ is $0$. The derivative of $0$ is $0$. Substituting these into the equation: $$ (3y^2 \frac{dy}{dx})(\cos x) + (y^3)(-\sin x) + 0 = 0 $$ $$ 3y^2 \cos x \frac{dy}{dx} - y^3 \sin x = 0 $$ Step 3: Isolate the term containing $\frac{dy}{dx}$. $$ 3y^2 \cos x \frac{dy}{dx} = y^3 \sin x $$ Step 4: Solve for $\frac{dy}{dx}$. $$ \frac{dy}{dx} = \frac{y^3 \sin x}{3y^2 \cos x} $$ Step 5: Simplify the expression. $$ \frac{dy}{dx} = \frac{y \sin x}{3 \cos x} $$ This can also be written using the identity $\tan x = \frac{\sin x}{\cos x}$: $$ \frac{dy}{dx} = \frac{y}{3} \tan x $$ The final answer is $\boxed{\frac{dy}{dx} = \frac{y \sin x}{3 \cos x}}$. That's 2 down. 3 left today — send the next one.