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You’ve uploaded a problem set asking to find the second derivative \(\frac{d^2y}{dx^2}\) implicitly in terms of \(x\) and \(y\).

Let's start with problem **53**:

### Given Equation:
\[
7xy + \sin x = 2
\]

We need to find the second derivative of \(y\) with respect to \(x\) by using implicit differentiation.

### Step 1: Differentiate implicitly with respect to \(x\).

#### First Derivative:
\[
\frac{d}{dx}(7xy) + \frac{d}{dx}(\sin x) = \frac{d}{dx}(2)
\]

Apply the product rule to \(7xy\):
\[
7 \left( y + x \frac{dy}{dx} \right) + \cos x = 0
\]

Simplify:
\[
7y + 7x \frac{dy}{dx} + \cos x = 0
\]

Solve for \(\frac{dy}{dx}\):
\[
7x \frac{dy}{dx} = -7y - \cos x
\]
\[
\frac{dy}{dx} = \frac{-7y - \cos x}{7x}
\]

#### First Derivative Result:
\[
\frac{dy}{dx} = \frac{-7y - \cos x}{7x}
\]

### Step 2: Differentiate again to find the second derivative.

We'll now take the derivative of \(\frac{dy}{dx}\) using the quotient rule.

\[
\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{-7y - \cos x}{7x} \right)
\]

Using the quotient rule: 
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]

Where:
- \( u = -7y - \cos x \)
- \( v = 7x \)

Now, differentiate \(u\) and \(v\):
- \(\frac{du}{dx} = -7 \frac{dy}{dx} + \sin x\)
- \(\frac{dv}{dx} = 7\)

Substitute into the quotient rule formula:
\[
\frac{d^2y}{dx^2} = \frac{7x \left( -7 \frac{dy}{dx} + \sin x \right) - (-7y - \cos x) \cdot 7}{(7x)^2}
\]

Simplify the expression. I'll calculate the rest now.The second derivative \(\frac{d^2y}{dx^2}\) for the equation \(7xy + \sin x = 2\) is:

\[
\frac{d^2y}{dx^2} = \frac{x \sin x + 7y + \cos x}{7x^2}
\]

Would you like more details on this process or any further clarifications?

Here are 5 related questions to deepen your understanding:
1. How does implicit differentiation work for non-linear equations?
2. What is the quotient rule and how is it applied in implicit differentiation?
3. Can implicit differentiation be used to find higher-order derivatives in similar problems?
4. How would the process change if \(y\) were a function of multiple variables?
5. What are some practical applications of second derivatives in physics and engineering?

**Tip**: When applying the product or quotient rule in implicit differentiation, always keep track of the chain rule for terms involving \(y\), as it allows you to relate the derivatives to \(\frac{dy}{dx}\).

Find the second derivative of 7xy + sin(x) = 2 implicitly in terms of x and y.

This problem demonstrates how to find the second derivative of the equation 7xy + sin(x) = 2 implicitly in terms of x and y using implicit differentiation. Key concepts include the product rule, chain rule, and quotient rule. Suitable for students in advanced high school calculus or early university courses.

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