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

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Answer 1 · 2017-03-26T00:44:57

#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)).

Product rule states that:

#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#

Therefore, using product rule, we get:

#d/dx (2+x)(2-3x) = (2+x)(-3) + (1)(2-3x)#

Now, all we have left to do is simplify.

#color(white)"XX" (2+x)(-3) + (1)(2-3x)#

# = (-6-3x) + (2-3x)#

# = -6x-4#

Final Answer