How do you differentiate # f(x)=(e^x+1)(x+1)sin(x)# using the product rule?

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Answer 1 · 2015-11-07T03:11:55

Treat two of the multiplicands as a single function, then apply the product rule twice.

The product rule states that #d/dxf(x)g(x) = f'(x)g(x) + f(x)g'(x)#

To apply it in this case, let's let
#f_1(x) = e^x + 1#
#f_2(x) = x+1#
#f_3(x) = sin(x)#
#f_4(x) = f_2(x)f_3(x) = (x+1)sin(x)#

Now we can write #f(x) = f_1(x)f_4(x)#
So, applying the product rule,
#f'(x) = f_1'(x)f_4(x) + f_1(x)f_4'(x)#

But #f_4(x) = f_2(x)f_3(x)#, so once again we apply the product rule
#f_4'(x) = f_2'(x)f_3(x) + f_2(x)f_3'(x)#

Then we can substitute to put #f'(x)# in terms of #f_1(x)#, #f_2(x)#, and #f_3(x)# and their derivatives.

#f'(x) = f_1'(x)(f_2(x)f_3(x)) + f_1(x)(f_2'(x)f_3(x) + f_2(x)f_3'(x))#

Now we calculate #f_1'(x)#, #f_2'(x)#, and #f_3'(x)#.
#f_1'(x) = e^x#
#f_2'(x) = 1#
#f_3'(x) = cos(x)#

Substituting this all in, we have
#f'(x) = e^x(x+1)sin(x) + (e^x+1)(1*sin(x) + (x+1)cos(x))#

How do you differentiate # f(x)=(e^x+1)*(x+1)*sin(x)# using the product rule?