How do you differentiate #f(x) =(2+x )( 2-3x)# using the product rule?
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#f(x) = (2 + x)(2 - 3x)#
let say #u = (2 + x)#, then #u' = 1# and #v =(2 - 3x)#, then #v' = -3#
#f'(x) = uv' + vu'#
#f'(x) = (2 + x)(-3) +(2 - 3x)(1)#
#f'(x) = -6 -3x +2 - 3x = -6x -4#
If we want to differentiate #f(x)=(2+x)(2-3x)# using the product rule we use the following #(f')(g)+(g')(f)#. Where #f# is your first term #(2+x)# and #g# is your second term #(2-3x)#. So, we take the #d/dx# of #f# which is 1 using the power rule #nx^(n-1)# keep in mind that the #d/dx# of a constant is zero, while #g# remains the same. At this point what we have is #1(2-3x)# now we take the #d/dx# of #g# which is #-3#, #f# remains the same.
After we have derived the equation we now have the following:
#1(2-3x)-3(2+x)#
Go ahead distribute and simplify:
#2-3x-6-3x=-6x-4#
Our final answer is #-6x-4#.
