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

Question

Emilia Aguilar · Accepted Answer

#f'(x)=16x(2x-3)^2#

Given:

#f(x)=h(x)*g(x)#

the Product Rule tells:

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

#h(x)=(2x+1)=>h'(x)=2+0=2# #g(x)=(2x-3)^3#

To find #g'(x)# you can use the Chain Rule:

#g(x)=[i(x)]^n=>g'(x)=n[i(x)]^(n-1)*i'(x)#

#:. g'(x)=3(2x-3)^2*(2+0)=2*3(2x-3)^2=# #=6(2x-3)^2#

Thus:

#f'(x)=2*(2x-3)^3+6(2x-3)^2*(2x+1)=2(2x-3)^2(2x-3+3(2x+1))=# #=2(2x-3)^2(2xcolor(green)cancel(-3)+6xcolor(green)cancel(+3))=2(2x-3)^2(8x)=# #=16x(2x-3)^2#

Christopher Agresta · Answer

undefined To differentiate $ f(x) = (2x + 1)(2x - 3)^3 $ using the product rule, apply the formula $ (uv)' = u'v + uv' $, where $ u = 2x + 1 $ and $ v = (2x - 3)^3 $. Then differentiate each term separately and apply the product rule formula:

$ u' = 2 $ (derivative of $ 2x + 1 $ with respect to $ x $)

$ v' = 3(2x - 3)^2 \cdot 2 $ (derivative of $ (2x - 3)^3 $ with respect to $ x $)

Now, apply the product rule:

$ f'(x) = (2)(2x - 3)^3 + (2x + 1)[3(2x - 3)^2 \cdot 2] $

Simplify the expression:

$ f'(x) = 2(2x - 3)^3 + (2x + 1)(6(2x - 3)^2) $

$ f'(x) = 2(2x - 3)^3 + 12(2x + 1)(2x - 3)^2 $

So, the derivative of $ f(x) $ with respect to $ x $ is $ f'(x) = 2(2x - 3)^3 + 12(2x + 1)(2x - 3)^2 $.

Paisley Johnson · Answer

undefined To differentiate the function $f(x) = (2x+1)(2x-3)^3$ using the product rule, you apply the formula which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

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

The derivative of $u(x)$ with respect to $x$ is $u'(x) = 2$, and the derivative of $v(x)$ with respect to $x$ is $v'(x) = 3(2x - 3)^2 \cdot 2$.

Using the product rule formula, the derivative of $f(x)$ is:

$$f'(x) = u'(x)v(x) + u(x)v'(x)$$

Substituting the values:

$$f'(x) = 2(2x - 3)^3 + (2x + 1) \cdot 3(2x - 3)^2 \cdot 2$$