https://file.aiquickdraw.com/mathbot/file/1729366851351lnzyjak3.jpg

The question in the image asks to find the equation of the tangent line to the curve:

\[
y = 5 \ln \left( x^5 e^{x^4} \right)
\]

at the point \( (1, 5) \), and the answer should be in the form \( y = mx + b \).

### Steps to Solve:

1. **Simplify the function**:
   \[
   y = 5 \ln \left( x^5 e^{x^4} \right)
   \]
   Using properties of logarithms:
   \[
   y = 5 \left[ \ln \left( x^5 \right) + \ln \left( e^{x^4} \right) \right]
   \]
   \[
   y = 5 \left( 5 \ln(x) + x^4 \right)
   \]
   \[
   y = 25 \ln(x) + 5x^4
   \]

2. **Differentiate to find the slope**:
   To find the slope \( m \), we need the derivative of \( y \) with respect to \( x \):
   \[
   \frac{dy}{dx} = \frac{d}{dx} \left( 25 \ln(x) + 5x^4 \right)
   \]
   Derivative of \( 25 \ln(x) \) is \( \frac{25}{x} \), and derivative of \( 5x^4 \) is \( 20x^3 \):
   \[
   \frac{dy}{dx} = \frac{25}{x} + 20x^3
   \]

3. **Evaluate the derivative at \( x = 1 \)**:
   \[
   \frac{dy}{dx} \bigg|_{x=1} = \frac{25}{1} + 20(1)^3 = 25 + 20 = 45
   \]
   So, the slope \( m = 45 \).

4. **Use the point-slope form**:
   The equation of the tangent line can be written as:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m = 45 \), \( (x_1, y_1) = (1, 5) \). Substituting:
   \[
   y - 5 = 45(x - 1)
   \]
   \[
   y - 5 = 45x - 45
   \]
   \[
   y = 45x - 40
   \]

### Final Answer:
The equation of the tangent line is:
\[
y = 45x - 40
\]

Would you like further details or have any other questions?

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Here are some related questions you might find useful:
1. How do logarithmic differentiation rules apply to other complex functions?
2. How can the product and chain rules be combined for more complex derivatives?
3. What is the geometric interpretation of a tangent line to a curve?
4. How does the derivative at a point determine the slope of the tangent line?
5. What are other ways to find tangent lines without differentiating directly?

**Tip:** Always simplify the logarithmic and exponential expressions before differentiating, as it can make the process much easier and less error-prone.

Find the equation of the tangent line to the curve y = 5ln(x^5 * e^(x^4)) at the point (1,5).

Learn how to find the equation of the tangent line to the curve y = 5ln(x^5 * e^(x^4)) at the point (1, 5). This problem involves simplifying logarithmic expressions, applying the product and chain rules, and using the derivative to find the slope.

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