How do you differentiate #f(x)=4x(2x+3)^2 # using the product rule?

Question

How do you differentiate #f(x)=4x(2x+3)^2  # using the product rule?

Paisley Johnson · Accepted Answer

#f'(x)=12(2x+3)(2x+1)#

The product rule states that for a function #f(x)=g(x)h(x)#, the derivative of the function is

#f'(x)=g'(x)h(x)+g(x)h'(x)#

Applying this to #f(x)=4x(2x+3)^2#, we see that

#f'(x)=color(red)(d/dx[4x])(2x+3)^2+4xcolor(blue)(d/dx[(2x+3)^2])#

These derivatives can be found individually and then plugged back in.

#color(red)(d/dx[4x]=4#

The chain rule is needed for the following.

#color(blue)(d/dx[(2x+3)^2])=2(2x+3)^1d/dx[2x+3]color(blue)(=4(2x+3)#

These demonstrate that

#f'(x)=4(2x+3)^2+4x(4(2x+3))#

Factor out a #(2x+3)# term.

#f'(x)=(2x+3)(4(2x+3)+16x)#

Simplify.

#f'(x)=12(2x+3)(2x+1)#

Additionally, this equals

#f'(x)=48x^2+96x+36#

Scarlett Allison · Answer

undefined To differentiate $ f(x) = 4x(2x + 3)^2 $ using the product rule:

1. Identify the functions $ u(x) $ and $ v(x) $ that are being multiplied together.
   Let $ u(x) = 4x $ and $ v(x) = (2x + 3)^2 $.

2. Differentiate each function with respect to $ x $.
   $ u'(x) = 4 $ (since the derivative of $ 4x $ is $ 4 $) and
   $ v'(x) = 2(2x + 3) \cdot 2 $ (by applying the chain rule and then multiplying by the derivative of the exponent).

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

4. Substitute the derivatives and the original functions into the formula:
   $ f'(x) = (4x) \cdot (2(2x + 3) \cdot 2) + ((2x + 3)^2) \cdot 4 $.

5. Simplify the expression:
   $ f'(x) = 4(2x + 3) \cdot (2(2x + 3)) + 4(2x + 3)^2 $.

6. Further simplify to get the final derivative:
   $ f'(x) = 8(2x + 3)^2 + 4(2x + 3)^2 $.

7. Combine like terms:
   $ f'(x) = 12(2x + 3)^2 $.

Thus, the derivative of $ f(x) = 4x(2x + 3)^2 $ with respect to $ x $ using the product rule is $ f'(x) = 12(2x + 3)^2 $.