Let [tex]3 \sin x + 6 \cos y = 3[/tex].

Find the first and second derivatives:

[tex]\[
\begin{aligned}
\frac{dy}{dx} &= \frac{1}{2} \left( \frac{\cos x}{\cos y} \right) \\
\frac{d^2 y}{dx^2} &= -\frac{1}{2} \left( \frac{\cos (x-y)}{(\sin y)^2} \right)
\end{aligned}
\][/tex]



Answer :

Sure, let's go through the problem step by step:

We start with the given equation:
[tex]\[ 3 \sin(x) + 6 \cos(y) = 3 \][/tex]

### Finding the First Derivative

To find [tex]\(\frac{dy}{dx}\)[/tex], we need to implicitly differentiate the given equation with respect to [tex]\(x\)[/tex].

1. Differentiate each term in the equation with respect to [tex]\(x\)[/tex]:

[tex]\[ \frac{d}{dx} [3 \sin(x)] + \frac{d}{dx} [6 \cos(y)] = \frac{d}{dx} [3] \][/tex]

2. Apply the chain rule to each term:
[tex]\[ 3 \cos(x) \cdot \frac{dx}{dx} + 6 \cdot (-\sin(y)) \cdot \frac{dy}{dx} = 0 \][/tex]

Simplify the derivatives:
[tex]\[ 3 \cos(x) - 6 \sin(y) \cdot \frac{dy}{dx} = 0 \][/tex]

3. Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ 6 \sin(y) \cdot \frac{dy}{dx} = 3 \cos(x) \][/tex]

[tex]\[ \frac{dy}{dx} = \frac{3 \cos(x)}{6 \sin(y)} \][/tex]

Simplify:
[tex]\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos(x)}{\sin(y)} \][/tex]

### Finding the Second Derivative

To find [tex]\(\frac{d^2y}{dx^2}\)[/tex], we need to differentiate [tex]\(\frac{dy}{dx}\)[/tex] again with respect to [tex]\(x\)[/tex].

Given:
[tex]\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos(x)}{\sin(y)} \][/tex]

1. Write [tex]\(\frac{dy}{dx}\)[/tex] in terms of [tex]\(u = \frac{\cos(x)}{\sin(y)}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{1}{2} u \][/tex]

2. Differentiate [tex]\(\frac{dy}{dx}\)[/tex] implicitly with respect to [tex]\(x\)[/tex]:

[tex]\[ \frac{d^2 y}{d x^2} = \frac{1}{2} \cdot \frac{d}{dx} \left( \frac{\cos(x)}{\sin(y)} \right) \][/tex]

3. Apply the quotient rule to [tex]\(\frac{\cos(x)}{\sin(y)}\)[/tex]:

[tex]\[ \frac{d}{dx} \left( \frac{\cos(x)}{\sin(y)} \right) = \frac{(\sin(y) \cdot (-\sin(x)) - \cos(x) \cdot \cos(y) \cdot \frac{dy}{dx})}{(\sin(y))^2} \][/tex]

Substitute [tex]\(\frac{dy}{dx} = \frac{1}{2} \frac{\cos(x)}{\sin(y)}\)[/tex] into the equation:

[tex]\[ = \frac{\sin(y) (-\sin(x)) - \cos(x) \cos(y) \left( \frac{1}{2} \cdot \frac{\cos(x)}{\sin(y)} \right)}{(\sin(y))^2} \][/tex]

Simplify:
[tex]\[ = \frac{-\sin(x) \sin(y) - \frac{1}{2} \cos(x)^2 \cdot \cos(y) / \sin(y)}{(\sin(y))^2} \][/tex]

[tex]\[ = \frac{-\sin(x) \sin(y) - \frac{1}{2} \cos(x) \cos(y) \cdot \cos(x) / \sin(y)}{(\sin(y))^2} \][/tex]

Finally, simplify the fraction:
[tex]\[ \frac{d^2 y}{dx^2} = -\frac{1}{2} \left( \frac{\cos(x - y)}{(\sin(y))^2} \right) \][/tex]

Thus, we have the final results:
- The given equation: [tex]\[ 3 \sin(x) + 6 \cos(y) = 3 \][/tex]
- The first derivative: [tex]\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos(x)}{\cos(y)} \][/tex]
- The second derivative: [tex]\[ \frac{d^2 y}{d x^2} = -\frac{1}{2} \left( \frac{\cos(x - y)}{\sin(y)^2} \right) \][/tex]