Given the function $f(x) = (x^2 + a)^5$, we are asked to find the coefficient of the term $x^5$ in the expansion of the derivative $f'(x)$, and it is known that this coefficient equals 960.

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

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

Step 2: Expand $(x^2 + a)^4$

To get the desired term $x^5$ after multiplying by $10x$, we need to expand $(x^2 + a)^4$.

Using the binomial expansion: $(x^2 + a)^4 = \sum_{k=0}^{4} \binom{4}{k} (x^2)^k a^{4-k}$ This gives: $(x^2 + a)^4 = \binom{4}{0} a^4 + \binom{4}{1} x^2 a^3 + \binom{4}{2} x^4 a^2 + \binom{4}{3} x^6 a + \binom{4}{4} x^8$ Substitute the binomial coefficients: $(x^2 + a)^4 = a^4 + 4x^2 a^3 + 6x^4 a^2 + 4x^6 a + x^8$

Step 3: Multiply by $10x$

Now, multiply $10x$ by each term of the expansion: $f'(x) = 10x \left( a^4 + 4x^2 a^3 + 6x^4 a^2 + 4x^6 a + x^8 \right)$ $f'(x) = 10x a^4 + 40x^3 a^3 + 60x^5 a^2 + 40x^7 a + 10x^9$

Step 4: Identify the coefficient of $x^5$

The term involving $x^5$ is $60x^5 a^2$. According to the problem, the coefficient of $x^5$ is 960, so: $60a^2 = 960$

Step 5: Solve for $a$

$a^2 = \frac{960}{60} = 16$ $a = \pm 4$

Thus, the possible values of $a$ are $a = 4$ and $a = -4$.

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Would you like more details on any step, or do you have any other questions?

Here are 5 related questions to deepen understanding:

1. What is the general form of the binomial expansion for $(x^n + a)^m$?
2. How does the chain rule apply when differentiating compositions of functions?
3. What happens to the coefficients of terms when multiplying polynomials by constants?
4. How would the problem change if the exponent in $(x^2 + a)^5$ was different, say 6?
5. How can you verify that the coefficient of $x^5$ in the expanded derivative is correct?

Tip: When expanding binomials, keep track of the powers and coefficients carefully to avoid mistakes in multi-step problems like this one.