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Sagot :
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]
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]
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