You are presenting the problem involving a second-order differential equation, and I believe you are asking for the second derivative of $y$ with respect to $x$, where the equation involves trigonometric functions. However, the expression seems unclear, as there's a mix of both $x$ and $y$, and you seem to be differentiating with respect to $x$.

Let’s break it down step-by-step:

The equation seems to be: $5 \sin(x) + 3 \cos(y) = 3$

Now, we'll differentiate both sides with respect to $x$, applying implicit differentiation because $y$ is a function of $x$.

Step 1: Differentiate $5 \sin(x)$

The derivative of $5 \sin(x)$ with respect to $x$ is: $\frac{d}{dx}[5 \sin(x)] = 5 \cos(x)$

Step 2: Differentiate $3 \cos(y)$

Now, using the chain rule, the derivative of $3 \cos(y)$ with respect to $x$ is: $\frac{d}{dx}[3 \cos(y)] = 3(-\sin(y)) \cdot \frac{dy}{dx} = -3 \sin(y) \frac{dy}{dx}$

Step 3: Differentiate the right side

The derivative of the constant 3 is 0: $\frac{d}{dx}[3] = 0$

Step 4: Combine the results

The equation after differentiation is: $5 \cos(x) - 3 \sin(y) \frac{dy}{dx} = 0$

Step 5: Solve for $\frac{dy}{dx}$

$- 3 \sin(y) \frac{dy}{dx} = -5 \cos(x)$ $\frac{dy}{dx} = \frac{5 \cos(x)}{3 \sin(y)}$

Step 6: Differentiate again to find $\frac{d^2y}{dx^2}$

Now, we need to differentiate $\frac{dy}{dx} = \frac{5 \cos(x)}{3 \sin(y)}$ with respect to $x$, applying the quotient rule and chain rule.

Using the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ where $u = 5 \cos(x)$ and $v = 3 \sin(y)$.

Differentiate $u = 5 \cos(x)$

$\frac{du}{dx} = -5 \sin(x)$

Differentiate $v = 3 \sin(y)$

Using the chain rule: $\frac{dv}{dx} = 3 \cos(y) \frac{dy}{dx}$ Substitute $\frac{dy}{dx} = \frac{5 \cos(x)}{3 \sin(y)}$ into this: $\frac{dv}{dx} = 3 \cos(y) \cdot \frac{5 \cos(x)}{3 \sin(y)} = \frac{5 \cos(x) \cos(y)}{\sin(y)}$

Now apply the quotient rule:

$\frac{d^2y}{dx^2} = \frac{3 \sin(y) (-5 \sin(x)) - 5 \cos(x) \left( \frac{5 \cos(x) \cos(y)}{\sin(y)} \right)}{(3 \sin(y))^2}$ Simplify this expression: $\frac{d^2y}{dx^2} = \frac{-15 \sin(y) \sin(x) - 25 \cos(x)^2 \cos(y)}{9 \sin(y)^2}$

This is the second derivative $\frac{d^2y}{dx^2}$.

Would you like more details, or do you have any questions?

Related Questions:

1. How can implicit differentiation be applied to more complex equations?
2. What is the role of the quotient rule in higher-order derivatives?
3. How do you solve second-order differential equations with mixed trigonometric terms?
4. Can the chain rule be used to simplify the differentiation process for functions of multiple variables?
5. How can we solve non-linear differential equations analytically?

Tip:

When using the quotient rule, carefully apply the chain rule when any variable depends on the one you're differentiating with respect to, especially in implicit differentiation cases.