Math Problem Statement

Evaluate the derivative of the function f(x) = √[(1 + e^(4x)) / (1 - e^(4x))]

Solution

We are tasked with finding the derivative of the function

f(x)=1+e4x1e4x.f(x) = \sqrt{\frac{1 + e^{4x}}{1 - e^{4x}}}.

To differentiate this function, we will apply the following steps:

Step 1: Simplify the expression

Let's rewrite the function in a more convenient form for differentiation:

f(x)=(1+e4x1e4x)1/2.f(x) = \left(\frac{1 + e^{4x}}{1 - e^{4x}}\right)^{1/2}.

Step 2: Apply the chain rule

We treat the function as a composition of two functions:

  • Outer function: g(u)=u1/2g(u) = u^{1/2}
  • Inner function: u(x)=1+e4x1e4xu(x) = \frac{1 + e^{4x}}{1 - e^{4x}}

First, differentiate the outer function g(u)g(u) with respect to uu:

g(u)=12u1/2.g'(u) = \frac{1}{2} u^{-1/2}.

Now differentiate the inner function u(x)u(x) with respect to xx using the quotient rule. The quotient rule states:

ddx(f(x)g(x))=f(x)g(x)f(x)g(x)g(x)2.\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}.

Here, f(x)=1+e4xf(x) = 1 + e^{4x} and g(x)=1e4xg(x) = 1 - e^{4x}.

  • The derivative of f(x)=1+e4xf(x) = 1 + e^{4x} is f(x)=4e4xf'(x) = 4e^{4x}.
  • The derivative of g(x)=1e4xg(x) = 1 - e^{4x} is g(x)=4e4xg'(x) = -4e^{4x}.

Now applying the quotient rule:

ddx(1+e4x1e4x)=(4e4x)(1e4x)(1+e4x)(4e4x)(1e4x)2.\frac{d}{dx} \left( \frac{1 + e^{4x}}{1 - e^{4x}} \right) = \frac{(4e^{4x})(1 - e^{4x}) - (1 + e^{4x})(-4e^{4x})}{(1 - e^{4x})^2}.

Simplifying the numerator:

(4e4x)(1e4x)(1+e4x)(4e4x)=4e4x4e8x+4e4x+4e8x=8e4x.(4e^{4x})(1 - e^{4x}) - (1 + e^{4x})(-4e^{4x}) = 4e^{4x} - 4e^{8x} + 4e^{4x} + 4e^{8x} = 8e^{4x}.

So, the derivative of the inner function is:

u(x)=8e4x(1e4x)2.u'(x) = \frac{8e^{4x}}{(1 - e^{4x})^2}.

Step 3: Combine the results

Now, apply the chain rule:

f(x)=g(u)u(x)=12(1+e4x1e4x)1/28e4x(1e4x)2.f'(x) = g'(u) \cdot u'(x) = \frac{1}{2} \left( \frac{1 + e^{4x}}{1 - e^{4x}} \right)^{-1/2} \cdot \frac{8e^{4x}}{(1 - e^{4x})^2}.

Simplifying:

f(x)=4e4x(1e4x)2(1+e4x1e4x)1/2.f'(x) = \frac{4e^{4x}}{(1 - e^{4x})^2} \cdot \left( \frac{1 + e^{4x}}{1 - e^{4x}} \right)^{-1/2}.

This is the derivative of the given function.

Would you like more details or explanations on any specific part of the process?

Here are some related questions to expand on this topic:

  1. How is the chain rule applied in composite functions?
  2. Can you explain the quotient rule in detail?
  3. How would you find the second derivative of the same function?
  4. What happens to the derivative as xx \to \infty?
  5. How would this function behave near x=0x = 0?

Tip: When differentiating composite functions, always identify the outer and inner functions first.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivative
Chain Rule
Quotient Rule
Exponential Functions

Formulas

Chain Rule: f'(g(x)) * g'(x)
Quotient Rule: d/dx[f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]^2
Exponential Function Derivative: d/dx[e^x] = e^x

Theorems

Chain Rule
Quotient Rule
Derivative of Exponential Functions

Suitable Grade Level

Undergraduate Calculus