How do you differentiate #g(y) =(2+x )( 2-3x) # using the product rule?

Question

How do you differentiate #g(y) =(2+x )( 2-3x)   # using the product rule?

William Abdalla · Accepted Answer

#d/dxg(x) = -6x-4#

(In the answer, I use h(x) where g(x) is traditionally used to avoid confusion since the question already defines g(x)).

According to the product rule, #d/dx f(x) * h(x) = f(x) * h'(x) + f'(x) * h(x)# In this case, #f(x) = 2+x#, and #h(x) = 2-3x#. If we differentiate each of these separately, we get that: #d/dx f(x) = d/dx (2+x) = 1# #d/dx h(x) = d/dx (2-3x) = -3# Thus, applying the product rule, we obtain: #d/dx (2+x)(2-3x) = (2+x)(-3) + (1)(2-3x)# Now, simplifying is the only thing left to do. #color(white)"XX" (2+x)(-3) + (1)(2-3x)# # = (-6-3x) + (2-3x)# # = -6x-4# Last Response

Adam Adkins · Answer

undefined To differentiate the function g(y) = (2 + x)(2 - 3x) using the product rule, you need to differentiate each factor separately and then apply the product rule.

First, differentiate the first factor (2 + x) with respect to x, which gives you 1. Then, differentiate the second factor (2 - 3x) with respect to x, which gives you -3.

Now, apply the product rule: 
g'(x) = (1)(2 - 3x) + (2 + x)(-3)
     = 2 - 3x - 3(2 + x)
     = 2 - 3x - 6 - 3x
     = -4 - 6x

So, the derivative of g(y) with respect to x is g'(x) = -4 - 6x.