2.6 Q-2

Question Statement

Find $f^{\prime}(\boldsymbol{x})$ for the following functions:

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

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

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

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

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

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

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

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

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Background and Explanation

To solve these derivative problems, we use the basic rules of differentiation, such as:

1. The Chain Rule: For a composite function, the derivative is the derivative of the outer function times the derivative of the inner function.
2. Product Rule: For the product of two functions, $(fg)' = f'g + fg$.
3. Quotient Rule: For the quotient of two functions, $\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$.
4. Logarithmic Differentiation: Useful when taking the derivative of a function with logarithms.

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Solution

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

We need to differentiate $e^{\sqrt{x} - 1}$ with respect to $x$. First, recognize the composition of functions: $e^{u}$ where $u = \sqrt{x} - 1$.

Step 1: Apply the chain rule:

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

Step 2: Differentiate $\sqrt{x} - 1$:

$$\frac{d}{dx}(\sqrt{x} - 1) = \frac{1}{2\sqrt{x}}$$

Step 3: Combine the results:

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

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

We apply the product rule because this is a product of two functions: $x^3$ and $e^{\frac{1}{x}}$.

Step 1: Differentiate using the product rule:

$$f'(x) = \frac{d}{dx}(x^3) \cdot e^{\frac{1}{x}} + x^3 \cdot \frac{d}{dx}(e^{\frac{1}{x}})$$

Step 2: Differentiate $x^3$:

$$\frac{d}{dx}(x^3) = 3x^2$$

Step 3: Differentiate $e^{\frac{1}{x}}$ using the chain rule:

$$\frac{d}{dx}\left(e^{\frac{1}{x}}\right) = e^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right)$$

Step 4: Combine the results:

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

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

We apply the product rule again, as this is a product of two functions: $e^x$ and $(1 + \ln x)$.

Step 1: Differentiate using the product rule:

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

Step 2: Simplify:

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

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

We use the quotient rule for this function.

Step 1: Apply the quotient rule:

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

Step 2: Differentiate:

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

Step 3: Simplify:

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

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

We apply the chain rule and logarithmic differentiation.

Step 1: Differentiate:

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

Step 2: Differentiate the terms inside the logarithm:

$$\frac{d}{dx}(e^x + e^{-x}) = e^x - e^{-x}$$

Step 3: Combine the results:

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

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

This is another function for which we will apply the quotient rule.

Step 1: Differentiate using the quotient rule:

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

Step 2: Differentiate the terms:

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

Step 3: Simplify:

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

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

We use the chain rule for this function.

Step 1: Differentiate using the chain rule:

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

Step 2: Differentiate the logarithm and the exponential terms:

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

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

We differentiate this using logarithmic differentiation.

Step 1: Simplify the expression:

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

Step 2: Differentiate:

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

Step 3: Simplify:

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

This is the hyperbolic tangent function:

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

Summary

• We applied **basic differentiation rules

** like the Chain Rule, Product Rule, and Quotient Rule to solve the problems.

• For composite functions, the Chain Rule was crucial.
• The Product Rule was used in problems with multiplication of functions, and the Quotient Rule helped solve fractional expressions.
• We simplified expressions, leveraging logarithmic properties where needed.