\frac{d^2y}{dx^2}\left[5sinx\:+\:3cosy\:=\:3\right]

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.

\frac{d^2y}{dx^2}\left[5sinx + 3cosy = 3\right]

Learn how to solve a second-order differential equation using implicit differentiation and the quotient rule. This problem involves the trigonometric equation 5sin(x) + 3cos(y) = 3, covering concepts like chain rule, quotient rule, and trigonometric functions. Suitable for advanced high school students (Grades 11-12).

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