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}\