2.6 Q-1

Question Statement

Find the derivatives of the following functions with respect to $x$:

1. $f(x) = e^{\sqrt{x} - 1}$
2. $f(x) = x^3 e^{\frac{1}{x}}$
3. $f(x) = e^x (1 + \ln x)$
4. $f(x) = \frac{e^x}{e^x + 1}$
5. $f(x) = \ln \left(e^x + e^{-x}\right)$
6. $f(x) = \frac{e^{a x} - e^{-a x}}{e^{a x} + e^{-a x}}$
7. $f(x) = \sqrt{\ln \left(e^{2 x} + e^{-2 x}\right)}$
8. $f(x) = \ln \left(\sqrt{e^{2 x} + e^{-2 x}}\right)$

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Background and Explanation

To solve these derivatives, we need to use the following concepts:

• Chain Rule: For composite functions, the chain rule is essential. It states that for $f(x) = g(h(x))$, the derivative is $f'(x) = g'(h(x)) \cdot h'(x)$.
• Product Rule: For functions that are products of two functions, such as $f(x) = g(x) \cdot h(x)$, the derivative is $f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$.
• Quotient Rule: For functions in the form $f(x) = \frac{g(x)}{h(x)}$, the derivative is $f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{h(x)^2}$.
• Logarithmic Differentiation: When dealing with logarithms, use the chain rule and properties of logarithms.

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Solution

i. $f(x) = e^{\sqrt{x} - 1}$

Apply the chain rule:

$$f'(x) = e^{\sqrt{x} - 1} \cdot \frac{d}{dx}(\sqrt{x} - 1)$$ $$= e^{\sqrt{x} - 1} \cdot \frac{1}{2\sqrt{x}}$$

Thus, the derivative is:

$$f'(x) = \frac{1}{2\sqrt{x}} e^{\sqrt{x} - 1}$$

ii. $f(x) = x^3 e^{\frac{1}{x}}$

Use the product rule:

$$f'(x) = e^{\frac{1}{x}} \cdot 3x^2 + x^3 \cdot e^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right)$$

Simplify:

$$f'(x) = 3x^2 e^{\frac{1}{x}} - x e^{\frac{1}{x}}$$ $$f'(x) = x e^{\frac{1}{x}} \left( 3x - 1 \right)$$

iii. $f(x) = e^x (1 + \ln x)$

Apply the product rule:

$$f'(x) = (1 + \ln x) \cdot e^x + e^x \cdot \frac{1}{x}$$

Simplify:

$$f'(x) = e^x \left( 1 + \ln x + \frac{1}{x} \right)$$

Thus, the derivative is:

$$f'(x) = \frac{(x(1 + \ln x) + 1) e^x}{x}$$

iv. $f(x) = \frac{e^x}{e^x + 1}$

Use the quotient rule:

$$f'(x) = \frac{\left( e^x + 1 \right) \cdot e^x - e^x \cdot \left( e^x \right)}{(e^x + 1)^2}$$

Simplify:

$$f'(x) = \frac{e^x + 2}{(e^x + 1)^2}$$

v. $f(x) = \ln \left(e^x + e^{-x}\right)$

Differentiate:

$$f'(x) = \frac{1}{e^x + e^{-x}} \cdot \left( e^x - e^{-x} \right)$$

Simplify:

$$f'(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

vi. $f(x) = \frac{e^{a x} - e^{-a x}}{e^{a x} + e^{-a x}}$

Differentiate using the quotient rule:

$$f'(x) = \frac{\left( e^{a x} + e^{-a x} \right) \cdot \left( a e^{a x} + a e^{-a x} \right) - \left( e^{a x} - e^{-a x} \right) \cdot \left( a e^{a x} \right)}{\left( e^{a x} + e^{-a x} \right)^2}$$

Simplify:

$$f'(x) = \frac{4a}{\left( e^{a x} + e^{-a x} \right)^2}$$

vii. $f(x) = \sqrt{\ln \left(e^{2x} + e^{-2x}\right)}$

Differentiate using the chain rule:

$$f'(x) = \frac{1}{2 \sqrt{\ln \left(e^{2 x} + e^{-2 x}\right)}} \cdot \frac{1}{e^{2x} + e^{-2x}} \cdot \left( 2 e^{2x} - 2 e^{-2x} \right)$$

Simplify:

$$f'(x) = \frac{e^{2x} - e^{-2x}}{\left( e^{2x} + e^{-2x} \right) \sqrt{\ln \left(e^{2x} + e^{-2x}\right)}}$$

viii. $f(x) = \ln \left( \sqrt{e^{2x} + e^{-2x}} \right)$

Differentiate using the chain rule:

$$f'(x) = \frac{1}{\sqrt{e^{2 x} + e^{-2 x}}} \cdot \frac{1}{2} \cdot \left( 2 e^{2x} - 2 e^{-2x} \right)$$

Simplify:

$$f'(x) = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}$$

Thus, this is the hyperbolic tangent:

$$f'(x) = \tanh(2x)$$

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx} e^{g(x)} = e^{g(x)} \cdot g'(x)$
• Product Rule: $\frac{d}{dx} \left( g(x) \cdot h(x) \right) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$
• Quotient Rule: $\frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{h(x)^2}$
• Logarithmic Differentiation: For $\frac{d}{dx} \ln f(x) = \frac{f'(x)}{f(x)}$

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Summary of Steps

1. For $f(x) = e^{\sqrt{x}-1}$, apply the chain rule to differentiate.
2. For $f(x) = x^3 e^{\frac{1}{x}}$, use the product rule and simplify.
3. For $f(x) = e^x (1 + \ln x)$, use the product rule and apply properties of logarithms.
4. For $f(x) = \frac{e^x}{e^x + 1}$, apply the quotient rule and simplify.
5. For $f(x) = \ln \left(e^x + e^{-x}\right)$, differentiate using basic differentiation rules.
6. For $f(x) = \frac{e^{a x} - e^{-a x}}{e^{a x} + e^{-a x}}$, apply the quotient rule and simplify.
7. For $f(x) = \sqrt{\ln \left(e^{2x} + e^{-2x}\right)}$, use the chain rule.
8. For $f(x) = \ln \left( \sqrt{e^{2x} + e^{-2x}} \right)$, differentiate using the chain rule and simplify.