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We need to find the equation of the tangent line to the curve:

\[
y = 3 \ln\left(x^2 e^{x^5}\right)
\]

at the point \( (1, 3) \).

### Step 1: Simplify the logarithmic expression
We can simplify the function inside the logarithm by using the logarithmic property:
\[
\ln(a \cdot b) = \ln(a) + \ln(b)
\]
Applying this to \( y = 3 \ln(x^2 e^{x^5}) \), we get:
\[
y = 3 \ln(x^2) + 3 \ln(e^{x^5})
\]
\[
y = 3 (2 \ln(x)) + 3 (x^5)
\]
\[
y = 6 \ln(x) + 3x^5
\]

### Step 2: Differentiate the simplified function
Now, we differentiate \( y = 6 \ln(x) + 3x^5 \) with respect to \( x \):
\[
\frac{dy}{dx} = \frac{d}{dx}(6 \ln(x)) + \frac{d}{dx}(3x^5)
\]
The derivative of \( 6 \ln(x) \) is:
\[
\frac{d}{dx}(6 \ln(x)) = \frac{6}{x}
\]
The derivative of \( 3x^5 \) is:
\[
\frac{d}{dx}(3x^5) = 15x^4
\]
Thus, the derivative \( \frac{dy}{dx} \) is:
\[
\frac{dy}{dx} = \frac{6}{x} + 15x^4
\]

### Step 3: Find the slope at \( x = 1 \)
To find the slope of the tangent line at \( x = 1 \), we substitute \( x = 1 \) into \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} \Big|_{x=1} = \frac{6}{1} + 15(1)^4 = 6 + 15 = 21
\]
So, the slope \( m \) of the tangent line at \( x = 1 \) is 21.

### Step 4: Use the point-slope form to find the equation of the tangent line
The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) = (1, 3) \) and \( m = 21 \). Substituting the values:
\[
y - 3 = 21(x - 1)
\]
Expanding the equation:
\[
y - 3 = 21x - 21
\]
\[
y = 21x - 18
\]

### Final Answer:
The equation of the tangent line is:
\[
y = 21x - 18
\]

Would you like more details or have any further questions?

Here are some related questions for deeper understanding:
1. How would you find the second derivative of \( y = 6 \ln(x) + 3x^5 \)?
2. How does the logarithmic property simplify complex logarithmic expressions?
3. Why is the point-slope form useful in finding tangent lines?
4. What if the point of tangency was different, say \( x = 2 \)?
5. How would the equation change if the original function was \( y = 3 \ln(x^2 e^{x^2}) \)?

**Tip:** Simplifying expressions before differentiating can make the process more manageable and reduce errors in complex functions.

Find the equation of the tangent line to the curve y = 3 ln(x^2 e^(x^5)) at the point (1, 3). Your answer must be an equation of the form y = mx + b.

This problem involves finding the equation of the tangent line to the curve y = 3 ln(x^2 e^(x^5)) at the point (1, 3). It covers differentiation, logarithmic properties, and finding slopes of tangent lines. Suitable for high school students in Grades 11-12.

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