Let's break down and solve the problem shown in the image.

Given:

The function is $\sin(y) = 2x^3 - 2$, and we need to find:

1. (a) $\frac{dy}{dx}$ using implicit differentiation.
2. (b) The slope of the curve at the point $(1, \pi)$.

Step (a): Use implicit differentiation to find $\frac{dy}{dx}$.

Starting with the given equation: $\sin(y) = 2x^3 - 2$

Differentiate both sides with respect to $x$, keeping in mind that $y$ is a function of $x$:

[ \frac{d}{dx}[\sin(y)] = \frac{d}{dx}[2x^3 - 2] ]

• The derivative of $\sin(y)$ with respect to $x$ is $\cos(y) \frac{dy}{dx}$, due to the chain rule.
• The derivative of $2x^3 - 2$ with respect to $x$ is $6x^2$.

Thus, the equation becomes: $\cos(y) \frac{dy}{dx} = 6x^2$

Solving for $\frac{dy}{dx}$, we get: $\frac{dy}{dx} = \frac{6x^2}{\cos(y)}$

Step (b): Find the slope at $(1, \pi)$.

At the point $(1, \pi)$, substitute $x = 1$ and $y = \pi$.

• For $x = 1$, we have $6x^2 = 6(1)^2 = 6$.
• For $y = \pi$, $\cos(\pi) = -1$.

Now, substitute these values into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{6}{-1} = -6$

So, the slope of the curve at the point $(1, \pi)$ is $-6$.

Would you like a further breakdown of these steps? Here are some related questions:

1. How does the chain rule work in implicit differentiation?
2. Why is the slope negative at the point $(1, \pi)$?
3. What other types of functions require implicit differentiation?
4. Can you differentiate $\sin(y)$ directly with respect to $x$?
5. How would the result change if we used another point?

Tip: Implicit differentiation is essential when you have equations involving both $x$ and $y$ without explicitly solving for one variable.