We are asked to find $y''$ (the second derivative of $y$) using implicit differentiation for the equation:

$\cos(y) + \sin(x) = 1$

Step 1: Differentiate both sides with respect to $x$

Differentiate both sides implicitly with respect to $x$:

[ \frac{d}{dx} \left[ \cos(y) + \sin(x) \right] = \frac{d}{dx} \left[ 1 \right] ]

On the left-hand side:

• The derivative of $\cos(y)$ with respect to $x$ requires the chain rule. The derivative of $\cos(y)$ is $-\sin(y)$, and then we multiply by $\frac{dy}{dx} = y'$, so $\frac{d}{dx}[\cos(y)] = -\sin(y) y'$.
• The derivative of $\sin(x)$ is $\cos(x)$.

On the right-hand side, the derivative of the constant 1 is 0.

Thus, the equation becomes:

$-\sin(y) y' + \cos(x) = 0$

Step 2: Solve for $y'$

Rearranging the equation to solve for $y'$:

$-\sin(y) y' = -\cos(x)$

$y' = \frac{\cos(x)}{\sin(y)}$

Step 3: Differentiate again to find $y''$

Now, differentiate $y' = \frac{\cos(x)}{\sin(y)}$ implicitly to find $y''$. Use the quotient rule to differentiate:

$y'' = \frac{(\sin(y) \cdot (-\sin(x)) - \cos(x) \cdot \cos(y) y')}{\sin^2(y)}$

Substitute $y' = \frac{\cos(x)}{\sin(y)}$ into this expression:

$y'' = \frac{\sin(y)(-\sin(x)) - \cos(x) \cos(y) \left( \frac{\cos(x)}{\sin(y)} \right)}{\sin^2(y)}$

Simplify the expression:

$y'' = \frac{-\sin(x) \sin(y) - \frac{\cos^2(x) \cos(y)}{\sin(y)}}{\sin^2(y)}$

Multiply both terms by $\sin(y)$ to combine:

$y'' = \frac{-\sin(x) \sin^2(y) - \cos^2(x) \cos(y)}{\sin^3(y)}$

Thus, the second derivative is:

$y'' = \frac{-\sin(x) \sin^2(y) - \cos^2(x) \cos(y)}{\sin^3(y)}$

Final result:

$y'' = \frac{-\sin(x) \sin^2(y) - \cos^2(x) \cos(y)}{\sin^3(y)}$

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

Here are 5 related questions:

1. How do you apply implicit differentiation to more complex equations?
2. What are some common uses of the chain rule in calculus?
3. How does the quotient rule work, and when do you use it?
4. How can implicit differentiation be used in real-world applications?
5. What is the significance of second derivatives in mathematics?

Tip: When performing implicit differentiation, always apply the chain rule when differentiating any term involving $y$, as it is implicitly a function of $x$.