To find $f''(x)$ for the function $f(x) = (x^2 + 7)^5$, we will use the chain rule and the power rule.

Step 1: Find the first derivative $f'(x)$

Using the chain rule: $f'(x) = 5(x^2 + 7)^4 \cdot \frac{d}{dx}(x^2 + 7)$ The derivative of $x^2 + 7$ is $2x$: $f'(x) = 5(x^2 + 7)^4 \cdot 2x = 10x(x^2 + 7)^4$

Step 2: Find the second derivative $f''(x)$

Now we differentiate $f'(x)$: $f''(x) = \frac{d}{dx}[10x(x^2 + 7)^4]$ Using the product rule: [ f''(x) = 10 \left[ (x^2 + 7)^4 + x \cdot \frac{d}{dx}[(x^2 + 7)^4] \right] ]

Now we find $\frac{d}{dx}[(x^2 + 7)^4]$ using the chain rule: $\frac{d}{dx}[(x^2 + 7)^4] = 4(x^2 + 7)^3 \cdot \frac{d}{dx}(x^2 + 7) = 4(x^2 + 7)^3 \cdot 2x = 8x(x^2 + 7)^3$

Substituting this back into our equation for $f''(x)$: $f''(x) = 10 \left[ (x^2 + 7)^4 + x \cdot 8x(x^2 + 7)^3 \right]$ $f''(x) = 10 \left[ (x^2 + 7)^4 + 8x^2(x^2 + 7)^3 \right]$ Factoring out $(x^2 + 7)^3$: $f''(x) = 10(x^2 + 7)^3 \left[ (x^2 + 7) + 8x^2 \right]$ $f''(x) = 10(x^2 + 7)^3 (9x^2 + 7)$

Thus, the final answer is: $f''(x) = 10(x^2 + 7)^3 (9x^2 + 7)$

Would you like further details on any part of the solution? Here are some related questions to consider:

1. What is the significance of the second derivative in calculus?
2. Can you explain the chain rule in more detail?
3. How does the product rule work for derivatives?
4. What are some applications of derivatives in real life?
5. How do you find higher-order derivatives?

Tip: Always check your work by differentiating your final result to ensure it aligns with your previous calculations!