Find $$ \frac{d y}{d x} $$ by implicit differentiation. $$ y \cos (x)=5 x^{2}+3 y^{2} $$ $$ \frac{d y}{d x}=\square $$

Question

UpStudy AI · Accepted Answer

We start with the equation
$$
y \cos x = 5x^2 + 3y^2.
$$

**Step 1. Differentiate both sides with respect to $$ x $$:**

Differentiate the left-hand side using the product rule:
$$
\frac{d}{dx}(y \cos x) = \frac{dy}{dx} \cos x - y \sin x.
$$

Differentiate the right-hand side:
$$
\frac{d}{dx}(5x^2 + 3y^2) = 10x + 6y\frac{dy}{dx}.
$$

**Step 2. Set the derivatives equal:**

$$
\frac{dy}{dx} \cos x - y \sin x = 10x + 6y\frac{dy}{dx}.
$$

**Step 3. Collect terms containing $$ \frac{dy}{dx} $$:**

Bring all terms with $$ \frac{dy}{dx} $$ to one side:
$$
\frac{dy}{dx} \cos x - 6y\frac{dy}{dx} = 10x + y \sin x.
$$

Factor out $$ \frac{dy}{dx} $$:
$$
\frac{dy}{dx} (\cos x - 6y) = 10x + y \sin x.
$$

**Step 4. Solve for $$ \frac{dy}{dx} $$:**

Divide both sides by $$ (\cos x - 6y) $$ (assuming $$ \cos x - 6y \neq 0 $$):
$$
\frac{dy}{dx} = \frac{10x + y \sin x}{\cos x - 6y}.
$$

Thus, the final answer is:
$$
\frac{dy}{dx} = \frac{10x + y \sin x}{\cos x - 6y}.
$$
$$ \frac{dy}{dx} = \frac{10x + y \sin x}{\cos x - 6y} $$

Find by implicit differentiation.