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The page contains various calculus problems. Let's go step by step to solve each problem. I'll break down the exercises according to the order given in the image.

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### **Beispiel 2: Tangentengleichung und Steigungswinkel**
We are tasked to find the equation of the tangent at a given point and the angle of inclination of the tangent with the horizontal axis.

Given:
- Function \( f(x) = x^3 - 5x \)
- Point \( P(-1|f(-1)) \)

Solution:
1. Derive the function:
   - \( f'(x) = 3x^2 - 5 \)
   - \( f'(-1) = 3(-1)^2 - 5 = 3 - 5 = -2 \)
   
2. The slope of the tangent is \( -2 \). Using the point-slope form of the tangent line:
   - \( y - f(-1) = -2(x + 1) \)
   - \( f(-1) = (-1)^3 - 5(-1) = -1 + 5 = 4 \)
   - So, the tangent equation becomes: \( y - 4 = -2(x + 1) \), simplified: \( y = -2x + 2 \)

3. Angle of inclination:
   - \( \tan(\alpha) = f'(-1) = -2 \)
   - \( \alpha = \arctan(-2) \approx -63.4^\circ \)

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### **Aufgaben (Exercises)**

#### **O1: Berechnen Sie die erste und zweite Ableitung (first and second derivatives) der Funktion f**

a) \( f(x) = 2\sin(x) + x^{-3} \)
   - First derivative: \( f'(x) = 2\cos(x) - 3x^{-4} \)
   - Second derivative: \( f''(x) = -2\sin(x) + 12x^{-5} \)

b) \( f(t) = -4t^3 - 3t \)
   - First derivative: \( f'(t) = -12t^2 - 3 \)
   - Second derivative: \( f''(t) = -24t \)

c) \( f(t) = \frac{t^4 - \sqrt{t}}{t} \)
   - First derivative: \( f'(t) = \frac{4t^3 - \frac{1}{2}t^{-1/2}}{t} \)

d) \( f(s) = \frac{\sqrt{s}}{5} - \frac{3}{s^5} \)
   - First derivative: \( f'(s) = \frac{1}{10\sqrt{s}} + 15s^{-6} \)
   - Second derivative: \( f''(s) = -\frac{1}{20s^{3/2}} - 90s^{-7} \)

e) \( f(x) = -\cos(x) + \frac{1}{x^2} \)
   - First derivative: \( f'(x) = \sin(x) - 2x^{-3} \)
   - Second derivative: \( f''(x) = \cos(x) + 6x^{-4} \)

f) \( f(x) = \frac{1}{\sqrt{3x}} + 3\cos(x) \)
   - First derivative: \( f'(x) = -\frac{1}{2}(3x)^{-3/2} + 3\sin(x) \)
   - Second derivative: \( f''(x) = \frac{3}{4}(3x)^{-5/2} + 3\cos(x) \)

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#### **O2: Berechnen Sie die erste Ableitung der Funktion f (First Derivatives)**

a) \( f(x) = (x + 4)^2 \)
   - First derivative: \( f'(x) = 2(x + 4) \)

b) \( f(s) = (2s - 4)(4 - s) \)
   - First derivative: Apply product rule
   - \( f'(s) = (2)(4 - s) + (2s - 4)(-1) = 8 - 2s - 2s + 4 = 12 - 4s \)

c) \( f(x) = x \cdot \left( x + \frac{1}{x^2} \right) \)
   - First derivative: Apply product and quotient rules
   - \( f'(x) = 1(x + \frac{1}{x^2}) + x\left(1 - \frac{2}{x^3}\

Solve calculus problems involving tangent line equations, first and second derivatives, and graph interpretations. Includes exercises on trigonometric functions, product and quotient rules, and function analysis. Suitable for students in Grades 10-12.

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