Let h(x) = (2x + 1)^5 (x^2 - 1)^3. Which expression is equivalent to h'(x)?

Step-by-step solution

1. 1

   Identify the outer structure as a product. $h$ is $u\cdot v$ with

   $u=(2x+1)^{5},\qquad v=\left(x^{2}-1\right)^{3},$

   so the product rule applies: $h'=u'v+uv'$. Options (C) and (D) contain only one term, so they are already wrong — they multiply the two derivatives instead of applying the product rule.

2. 2

   Differentiate u with the chain rule. Bring down the power, reduce it by one, and multiply by the derivative of the inside:

   $u'=5(2x+1)^{4}\cdot\frac{d}{dx}(2x+1)=5(2x+1)^{4}\cdot 2=10(2x+1)^{4}.$

   The factor $2$ is what distinguishes option (B) from option (A).

3. 3

   Differentiate v with the chain rule.

   $v'=3\left(x^{2}-1\right)^{2}\cdot\frac{d}{dx}\left(x^{2}-1\right)=3\left(x^{2}-1\right)^{2}\cdot 2x=6x\left(x^{2}-1\right)^{2}.$

4. 4

   Assemble the product rule.

   $h'(x)=10(2x+1)^{4}\left(x^{2}-1\right)^{3}+6x(2x+1)^{5}\left(x^{2}-1\right)^{2},$

   which is exactly option (B).

5. 5

   See why (A) is the tempting wrong answer. Option (A) is what you get by forgetting both inner derivatives — it uses $5$ and $3$ instead of $10$ and $6x$. Whenever the inside of a bracket is anything more complicated than $x$, that inner derivative must appear.

6. 6

   Verify symbolically. Differentiating the original expression and subtracting option (B) simplifies to exactly $0$, confirming the choice. As an extra check, the common factor $(2x+1)^4(x^2-1)^2$ can be pulled out to give $h'(x)=(2x+1)^4(x^2-1)^2\left[10(x^2-1)+6x(2x+1)\right]=(2x+1)^4(x^2-1)^2\left(22x^2+6x-10\right)$.