Derivatives of Exponential and Logarithmic Functions: Practice Problems

Problem: Find the derivatives of the following functions.

We will solve each function step by step using the rules of differentiation, particularly focusing on the derivatives of exponential and logarithmic functions. The key formulas we will use are:

1. Derivative of \( e^x \): \( \frac{d}{dx} e^x = e^x \)
2. Derivative of \( \ln(x) \): \( \frac{d}{dx} \ln(x) = \frac{1}{x} \)
3. Chain Rule: If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)
4. Product Rule: If \( y = u(x)v(x) \), then \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \)
5. Quotient Rule: If \( y = \frac{u(x)}{v(x)} \), then \( \frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \)

---

#### 1. \( f(x) = e^{2x} \)

Using the chain rule:
\[
f'(x) = \frac{d}{dx} e^{2x} = e^{2x} \cdot \frac{d}{dx}(2x) = e^{2x} \cdot 2 = 2e^{2x}
\]

#### 2. \( f(x) = \ln(5x) \)

Using the chain rule:
\[
f'(x) = \frac{d}{dx} \ln(5x) = \frac{1}{5x} \cdot \frac{d}{dx}(5x) = \frac{1}{5x} \cdot 5 = \frac{1}{x}
\]

#### 3. \( f(x) = x^3 - 5x + e^{x+3} \)

Differentiate each term separately:
\[
f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(5x) + \frac{d}{dx}(e^{x+3})
\]
\[
f'(x) = 3x^2 - 5 + e^{x+3} \cdot \frac{d}{dx}(x+3) = 3x^2 - 5 + e^{x+3} \cdot 1 = 3x^2 - 5 + e^{x+3}
\]

#### 4. \( f(x) = \ln(x^2) \)

Using the chain rule:
\[
f'(x) = \frac{d}{dx} \ln(x^2) = \frac{1}{x^2} \cdot \frac{d}{dx}(x^2) = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x}
\]

#### 5. \( f(x) = x^4 \ln(x) - 4x^4 \)

Differentiate each term separately using the product rule for the first term:
\[
f'(x) = \frac{d}{dx}(x^4 \ln(x)) - \frac{d}{dx}(4x^4)
\]
For \( x^4 \ln(x) \):
\[
\frac{d}{dx}(x^4 \ln(x)) = \frac{d}{dx}(x^4) \cdot \ln(x) + x^4 \cdot \frac{d}{dx}(\ln(x)) = 4x^3 \ln(x) + x^4 \cdot \frac{1}{x} = 4x^3 \ln(x) + x^3
\]
For \( 4x^4 \):
\[
\frac{d}{dx}(4x^4) = 16x^3
\]
Thus:
\[
f'(x) = 4x^3 \ln(x) + x^3 - 16x^3 = 4x^3 \ln(x) - 15x^3
\]

#### 6. \( f(x) = \ln\left(\frac{x+4}{x^3}\right) \)

Use the properties of logarithms:
\[
f(x) = \ln(x+4) - \ln(x^3) = \ln(x+4) - 3\ln(x)
\]
Differentiate each term:
\[
f'(x) = \frac{d}{dx} \ln(x+4) - \frac{d}{dx} (3\ln(x)) = \frac{1}{x+4} \cdot \frac{d}{dx}(x+4) - 3 \cdot \frac{1}{x}
\]
\[
f'(x) = \frac{1}{x+4} \cdot 1 - \frac{3}{x} = \frac{1}{x+4} - \frac{3}{x}
\]

#### 7. \( f(x) = x^3 e^x + \ln(1) \)

Note that \( \ln(1) = 0 \), so the function simplifies to:
\[
f(x) = x^3 e^x
\]
Use the product rule:
\[
f'(x) = \frac{d}{dx}(x^3) \cdot e^x + x^3 \cdot \frac{d}{dx}(e^x) = 3x^2 e^x + x^3 e^x = e^x (3x^2 + x^3)
\]

#### 8. \( f(x) = \ln(e^x) \)

Simplify using the property of logarithms:
\[
f(x) = \ln(e^x) = x
\]
Thus:
\[
f'(x) = 1
\]

#### 9. \( f(x) = \ln(e^x - 3x) \)

Use the chain rule:
\[
f'(x) = \frac{d}{dx} \ln(e^x - 3x) = \frac{1}{e^x - 3x} \cdot \frac{d}{dx}(e^x - 3x)
\]
\[
f'(x) = \frac{1}{e^x - 3x} \cdot (e^x - 3) = \frac{e^x - 3}{e^x - 3x}
\]

#### 10. \( f(x) = \ln(x \sqrt{x+2}) \)

Use the properties of logarithms:
\[
f(x) = \ln(x) + \ln(\sqrt{x+2}) = \ln(x) + \frac{1}{2} \ln(x+2)
\]
Differentiate each term:
\[
f'(x) = \frac{d}{dx} \ln(x) + \frac{d}{dx} \left( \frac{1}{2} \ln(x+2) \right) = \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x+2} \cdot \frac{d}{dx}(x+2)
\]
\[
f'(x) = \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x+2} \cdot 1 = \frac{1}{x} + \frac{1}{2(x+2)}
\]

#### 11. \( f(x) = 1 - e^{-0.087x} \)

Differentiate each term:
\[
f'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(e^{-0.087x}) = 0 - e^{-0.087x} \cdot \frac{d}{dx}(-0.087x)
\]
\[
f'(x) = -e^{-0.087x} \cdot (-0.087) = 0.087 e^{-0.087x}
\]

#### 12. \( f(x) = \sqrt{e^{2x} + 3x - 1} \)

Rewrite as a power:
\[
f(x) = (e^{2x} + 3x - 1)^{1/2}
\]
Use the chain rule:
\[
f'(x) = \frac{1}{2} (e^{2x} + 3x - 1)^{-1/2} \cdot \frac{d}{dx}(e^{2x} + 3x - 1)
\]
\[
f'(x) = \frac{1}{2} (e^{2x} + 3x - 1)^{-1/2} \cdot (2e^{2x} + 3) = \frac{2e^{2x} + 3}{2\sqrt{e^{2x} + 3x - 1}}
\]

#### 13. \( f(x) = \ln\left(\frac{x^4 + 2x^3}{x^2}\right) \)

Simplify the argument of the logarithm:
\[
f(x) = \ln\left(\frac{x^4 + 2x^3}{x^2}\right) = \ln(x^2 + 2x)
\]
Differentiate using the chain rule:
\[
f'(x) = \frac{1}{x^2 + 2x} \cdot \frac{d}{dx}(x^2 + 2x) = \frac{1}{x^2 + 2x} \cdot (2x + 2) = \frac{2x + 2}{x^2 + 2x} = \frac{2(x + 1)}{x(x + 2)}
\]

#### 14. \( f(x) = \ln\left(\frac{e^{2x} + 1}{2x}\right) \)

Use the properties of logarithms:
\[
f(x) = \ln(e^{2x} + 1) - \ln(2x)
\]
Differentiate each term:
\[
f'(x) = \frac{d}{dx} \ln(e^{2x} + 1) - \frac{d}{dx} \ln(2x) = \frac{1}{e^{2x} + 1} \cdot \frac{d}{dx}(e^{2x} + 1) - \frac{1}{2x} \cdot \frac{d}{dx}(2x)
\]
\[
f'(x) = \frac{1}{e^{2x} + 1} \cdot 2e^{2x} - \frac{1}{2x} \cdot 2 = \frac{2e^{2x}}{e^{2x} + 1} - \frac{1}{x}
\]

#### 15. \( f(x) = x^2 e^{(2x-5)} \)

Use the product rule:
\[
f'(x) = \frac{d}{dx}(x^2) \cdot e^{(2x-5)} + x^2 \cdot \frac{d}{dx}(e^{(2x-5)})
\]
\[
f'(x) = 2x e^{(2x-5)} + x^2 \cdot e^{(2x-5)} \cdot \frac{d}{dx}(2x-5) = 2x e^{(2x-5)} + x^2 e^{(2x-5)} \cdot 2
\]
\[
f'(x) = 2x e^{(2x-5)} + 2x^2 e^{(2x-5)} = 2e^{(2x-5)} (x + x^2)
\]

#### 16. \( f(x) = 2e^{\left(\frac{x^2}{2} - x \ln(x)\right)} \)

Use the chain rule:
\[
f'(x) = 2 \cdot e^{\left(\frac{x^2}{2} - x \ln(x)\right)} \cdot \frac{d}{dx}\left(\frac{x^2}{2} - x \ln(x)\right)
\]
Differentiate the exponent:
\[
\frac{d}{dx}\left(\frac{x^2}{2} - x \ln(x)\right) = \frac{d}{dx}\left(\frac{x^2}{2}\right) - \frac{d}{dx}(x \ln(x))
\]
\[
\frac{d}{dx}\left(\frac{x^2}{2}\right) = x
\]
\[
\frac{d}{dx}(x \ln(x)) = \frac{d}{dx}(x) \cdot \ln(x) + x \cdot \frac{d}{dx}(\ln(x)) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1
\]
Thus:
\[
\frac{d}{dx}\left(\frac{x^2}{2} - x \ln(x)\right) = x - (\ln(x) + 1) = x - \ln(x) - 1
\]
So:
\[
f'(x) = 2e^{\left(\frac{x^2}{2} - x \ln(x)\right)} \cdot (x - \ln(x) - 1)
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad 2e^{2x} \\
2. & \quad \frac{1}{x} \\
3. & \quad 3x^2 - 5 + e^{x+3} \\
4. & \quad \frac{2}{x} \\
5. & \quad 4x^3 \ln(x) - 15x^3 \\
6. & \quad \frac{e^x - 3}{e^x - 3x} \\
7. & \quad e^x (3x^2 + x^3) \\
8. & \quad 1 \\
9. & \quad \frac{e^x - 3}{e^x - 3x} \\
10. & \quad \frac{1}{x} + \frac{1}{2(x+2)} \\
11. & \quad 0.087 e^{-0.087x} \\
12. & \quad \frac{2e^{2x} + 3}{2\sqrt{e^{2x} + 3x - 1}} \\
13. & \quad \frac{2(x + 1)}{x(x + 2)} \\
14. & \quad \frac{2e^{2x}}{e^{2x} + 1} - \frac{1}{x} \\
15. & \quad 2e^{(2x-5)} (x + x^2) \\
16. & \quad 2e^{\left(\frac{x^2}{2} - x \ln(x)\right)} (x - \ln(x) - 1)
\end{aligned}
}
\]