Math Problem Statement

Find the derivative of the function y = sqrt(5 + 8e^(4x)).

Solution

To find the derivative of the given function y=5+8e4xy = \sqrt{5 + 8e^{4x}}, we proceed step by step:

Step 1: Rewrite the square root

Rewrite the square root using the fractional exponent: y=(5+8e4x)1/2.y = \left(5 + 8e^{4x}\right)^{1/2}.

Step 2: Differentiate using the chain rule

The derivative of y=(5+8e4x)1/2y = \left(5 + 8e^{4x}\right)^{1/2} with respect to xx is: y=12(5+8e4x)1/2ddx(5+8e4x).y' = \frac{1}{2} \left(5 + 8e^{4x}\right)^{-1/2} \cdot \frac{d}{dx}(5 + 8e^{4x}).

Step 3: Differentiate the inner function

The derivative of 5+8e4x5 + 8e^{4x} is: ddx(5+8e4x)=0+8ddx(e4x)=84e4x=32e4x.\frac{d}{dx}(5 + 8e^{4x}) = 0 + 8 \cdot \frac{d}{dx}(e^{4x}) = 8 \cdot 4e^{4x} = 32e^{4x}.

Step 4: Combine the results

Substitute ddx(5+8e4x)=32e4x\frac{d}{dx}(5 + 8e^{4x}) = 32e^{4x} into the chain rule result: y=12(5+8e4x)1/232e4x.y' = \frac{1}{2} \left(5 + 8e^{4x}\right)^{-1/2} \cdot 32e^{4x}.

Step 5: Simplify

Simplify the expression: y=32e4x25+8e4x.y' = \frac{32e^{4x}}{2\sqrt{5 + 8e^{4x}}}.

y=16e4x5+8e4x.y' = \frac{16e^{4x}}{\sqrt{5 + 8e^{4x}}}.

Final Answer:

y=16e4x5+8e4x.y' = \frac{16e^{4x}}{\sqrt{5 + 8e^{4x}}}.

Let me know if you would like a detailed explanation or further assistance! Here are some related questions to expand your understanding:

  1. How do we apply the chain rule to functions with nested exponents?
  2. What happens to the derivative when the exponent is negative instead of fractional?
  3. How would the derivative change if the base of the exponential term was a different number?
  4. How can we graph the derivative yy' to understand its behavior?
  5. What are the steps to confirm the derivative using implicit differentiation?

Tip: Always simplify your results at each step for clarity and correctness!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Exponential Functions

Formulas

Derivative of a square root: d/dx[√u] = (1 / 2√u) * du/dx
Chain Rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)
Derivative of e^(kx): d/dx[e^(kx)] = ke^(kx)

Theorems

Chain Rule

Suitable Grade Level

Undergraduate Calculus or Advanced High School Math