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 $\frac{dy}{dx}$.