To find the equation of the tangent to the graph of $f(x) = \frac{e^{4x}}{4x - 3}$, follow these steps:

General Form of Tangent Line

The equation of the tangent line at $x = a$ is: $y = f'(a)(x - a) + f(a)$

1. Compute $f'(x)$

Use the quotient rule to differentiate $f(x)$. The quotient rule states: $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$ Here:

• $u = e^{4x}$, so $u' = 4e^{4x}$
• $v = 4x - 3$, so $v' = 4$

Differentiating: $f'(x) = \frac{(4e^{4x})(4x - 3) - (e^{4x})(4)}{(4x - 3)^2}$ Simplify: $f'(x) = \frac{4e^{4x}(4x - 3) - 4e^{4x}}{(4x - 3)^2}$ Factor $4e^{4x}$ from the numerator: $f'(x) = \frac{4e^{4x} \left( (4x - 3) - 1 \right)}{(4x - 3)^2}$ $f'(x) = \frac{4e^{4x}(4x - 4)}{(4x - 3)^2}$ $f'(x) = \frac{16e^{4x}(x - 1)}{(4x - 3)^2}$

2. Evaluate $f(a)$ and $f'(a)$

At a specific point $x = a$, compute:

• $f(a) = \frac{e^{4a}}{4a - 3}$
• $f'(a) = \frac{16e^{4a}(a - 1)}{(4a - 3)^2}$

3. Write the Tangent Line Equation

Substitute $f(a)$ and $f'(a)$ into the tangent line equation: $y = f'(a)(x - a) + f(a)$ $y = \frac{16e^{4a}(a - 1)}{(4a - 3)^2}(x - a) + \frac{e^{4a}}{4a - 3}$

This is the equation of the tangent line.

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Would you like me to evaluate this at a specific point or simplify further?

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5 Questions to Expand Understanding

1. What is the importance of the tangent line in understanding the behavior of a function?
2. How does the quotient rule apply to finding derivatives of rational functions?
3. Why is it necessary to compute both $f(a)$ and $f'(a)$ for the tangent line equation?
4. What are the challenges of finding tangent lines for exponential functions?
5. How does the derivative of $e^{4x}$ differ from polynomial derivatives?

Tip

Always double-check derivative steps, especially when applying the product, quotient, or chain rules together.