Sure! Let's go through each question step-by-step.
1. Derivative of \( y = 2 \sin^3(x) - 4 \cos^2(x) \)
To find the derivative, we'll use the chain rule and the power rule.
First, differentiate each term:
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For \( 2 \sin^3(x) \):
- Let \( u = \sin(x) \), so \( y = 2u^3 \)
- \( \frac{dy}{du} = 6u^2 \)
- Using the chain rule, \( \frac{du}{dx} = \cos(x) \)
- So, \( \frac{dy}{dx} = 6 \sin^2(x) \cos(x) \)
-
For \( -4 \cos^2(x) \):
- Let \( v = \cos(x) \), so \( y = -4v^2 \)
- \( \frac{dy}{dv} = -8v \)
- Using the chain rule, \( \frac{dv}{dx} = -\sin(x) \)
- So, \( \frac{dy}{dx} = -8 \cos(x)(-\sin(x)) = 8 \cos(x) \sin(x) \)
Now, combine the derivatives: \[ \frac{dy}{dx} = 6 \sin^2(x) \cos(x) + 8 \cos(x) \sin(x) = \cos(x) \left( 6 \sin^2(x) + 8 \sin(x) \right) \]
2. Derivative of \( y = x e^x \)
To differentiate this, we'll use the product rule, which states \( (uv)' = u'v + uv' \).
Let \( u = x \) and \( v = e^x \):
- \( u' = 1 \)
- \( v' = e^x \)
Using the product rule: \[ \frac{dy}{dx} = u'v + uv' = 1 \cdot e^x + x \cdot e^x = e^x + xe^x = e^x(1 + x) \]
3. Derivative of \( y = 7 \ln(x^3 + 2) \)
We'll use the chain rule again. The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
Let \( u = x^3 + 2 \):
- \( \frac{du}{dx} = 3x^2 \)
So applying the chain rule: \[ \frac{dy}{dx} = 7 \cdot \frac{1}{x^3 + 2} \cdot 3x^2 = \frac{21x^2}{x^3 + 2} \]
4. Maximizing the Crop Yield
Let \( n \) be the number of trees per hectare. The number of apples per tree decreases by 4 for each additional tree planted. The yield can be expressed as: \[ \text{Yield} = n(400 - 4(n - 80)) \]
Expanding this: \[ \text{Yield} = n(400 - 4n + 320) = n(720 - 4n) \]
The yield function is a quadratic equation given by: \[ A = 720n - 4n^2 \] This is a downward-opening parabola. The maximum occurs at the vertex, where \( n = -\frac{b}{2a} = -\frac{720}{2 \cdot -4} = 90 \).
The maximum yield occurs at 90 trees per hectare.
5. Finding Maximum and Minimum Displacements
Given \( s(t) = \sin(t) - \cos(t) \).
To find the maximum and minimum, we need to differentiate and set the derivative to zero.
\[ \frac{ds}{dt} = \cos(t) + \sin(t) \] Setting this equal to zero: \[ \cos(t) + \sin(t) = 0 \Rightarrow \sin(t) = -\cos(t) \Rightarrow \tan(t) = -1 \]
This occurs for \( t = \frac{3\pi}{4} + k\pi \) within \( [0, 2\pi] \):
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At \( t = \frac{3\pi}{4} \): \[ s\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \]
-
At \( t = \frac{7\pi}{4} \): \[ s\left(\frac{7\pi}{4}\right) = \sin\left(\frac{7\pi}{4}\right) - \cos\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 \]
So the maximum is \( \sqrt{2} \) meters and minimum is \( 0 \) meters.
6. Maximum Area Rectangle with Given Perimeter
Let the sides of the rectangle be \( x \) and \( y \). The perimeter \( P = 2x + 2y \) is given, so: \[ y = \frac{P}{2} - x \]
To find the area \( A \): \[ A = xy = x\left(\frac{P}{2} - x\right) = \frac{P}{2}x - x^2 \]
This is a quadratic equation in the form \( A = -x^2 + \frac{P}{2}x \), which opens downward (as coefficient of \( x^2 \) is negative), so the maximum occurs at the vertex: \[ x = -\frac{b}{2a} = -\frac{\frac{P}{2}}{2 \cdot -1} = \frac{P}{4} \] Then, \[ y = \frac{P}{4} \]
Thus, both sides are equal, meaning the rectangle is a square.
Conclusion: A rectangle with the given perimeter has maximum area when it is a square.
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