How do you use the product rule to differentiate #f(x)=(x^2-3x+1)(3x+2)#?

Question

Elijah Abram · Accepted Answer

The product rule states that:

#(d(g(x)*h(x)))/dx = (dg(x))/dx*h(x) + g(x)*(dh(x))/dx# Pose: #g(x) = (x^2-3x+1) and h(x) =3x+2# #(df(x))/dx = (d(x^2-3x+1))/dx*(3x+2)+(x^2-3x+1)*(d(3x+2))/dx = (2x-3)(3x+2)+3(x^2-3x+1) = # # = 6x^2+4x-9x-6+3x^2-9x+3= 9x^2-14x-3#

Charles Anctil · Answer

undefined To differentiate $f(x) = (x^2 - 3x + 1)(3x + 2)$ using the product rule:

1. Identify the functions $u(x)$ and $v(x)$.
   Let $u(x) = x^2 - 3x + 1$ and $v(x) = 3x + 2$.

2. Apply the product rule formula:
   $$f'(x) = u'(x)v(x) + u(x)v'(x)$$

3. Find the derivatives of $u(x)$ and $v(x)$:
   $$u'(x) = \frac{d}{dx}(x^2 - 3x + 1) = 2x - 3$$
   $$v'(x) = \frac{d}{dx}(3x + 2) = 3$$

4. Substitute $u'(x)$, $v(x)$, $u(x)$, and $v'(x)$ into the product rule formula:
   $$f'(x) = (2x - 3)(3x + 2) + (x^2 - 3x + 1)(3)$$

5. Expand and simplify the expression:
   $$f'(x) = 6x^2 + 4x - 9x - 6 + 3x^2 - 9x + 3$$

6. Combine like terms:
   $$f'(x) = 9x^2 - 14x - 3$$

Therefore, the derivative of $f(x) = (x^2 - 3x + 1)(3x + 2)$ with respect to $x$ is $f'(x) = 9x^2 - 14x - 3$.