https://file.aiquickdraw.com/mathbot/file/1726627172408y0vbgouy.jpg

To find the second derivative \( y'' \) of the function \( y = x(3x + 1)^5 \), we first need to find the first derivative \( y' \) using the product rule, which states that if \( y = uv \), then \( y' = u'v + uv' \).

Let's define \( u = x \) and \( v = (3x + 1)^5 \). Then:
- \( u' = 1 \)
- \( v = (3x + 1)^5 \)

To find \( v' \), we use the chain rule. If \( v = f(g(x)) \), then \( v' = f'(g(x))g'(x) \). Here, \( f(u) = u^5 \) and \( g(x) = 3x + 1 \), so:
- \( f'(u) = 5u^4 \)
- \( g'(x) = 3 \)

Thus:
\[ v' = 5(3x + 1)^4 \cdot 3 = 15(3x + 1)^4 \]

Now, applying the product rule:
\[ y' = u'v + uv' = 1 \cdot (3x + 1)^5 + x \cdot 15(3x + 1)^4 = (3x + 1)^5 + 15x(3x + 1)^4 \]

Next, we find \( y'' \) by differentiating \( y' \) again. We need to apply the product and chain rules to each term:

1. Differentiating \( (3x + 1)^5 \):
\[ \frac{d}{dx}[(3x + 1)^5] = 5(3x + 1)^4 \cdot 3 = 15(3x + 1)^4 \]

2. Differentiating \( 15x(3x + 1)^4 \) using the product rule:
\[ \frac{d}{dx}[15x(3x + 1)^4] = 15 \cdot \left[ x \cdot \frac{d}{dx}[(3x + 1)^4] + (3x + 1)^4 \cdot \frac{d}{dx}[x] \right] \]
\[ = 15 \left[ x \cdot 4(3x + 1)^3 \cdot 3 + (3x + 1)^4 \right] \]
\[ = 180x(3x + 1)^3 + 15(3x + 1)^4 \]

Adding these terms together, we get:
\[ y'' = 15(3x + 1)^4 + 180x(3x + 1)^3 + 15(3x + 1)^4 \]
\[ y'' = 30(3x + 1)^4 + 180x(3x + 1)^3 \]

Let me know if you need further details or have any questions about these steps.

Questions to consider:
1. Would you like to learn more about how to use the chain rule in differentiating composite functions?
2. How does the product rule simplify the differentiation of products of functions?
3. What are the implications of having higher powers of a function within a product when differentiating?
4. Can you identify scenarios in physics or engineering where this kind of differentiation might be particularly useful?
5. Are you interested in exploring how third or higher derivatives of functions are calculated?

Tip: To visualize the effect of the second derivative on a function, consider its graph. The second derivative tells you about the curvature of the graph: where it is concave up or down. This can help in understanding inflection points and the behavior of the function's graph in general.

Learn how to calculate the second derivative of y = x(3x + 1)^5 using the product and chain rules. A step-by-step breakdown covers key differentiation concepts suitable for advanced high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation