How do you differentiate #g(x) = sqrt(x-3)e^(4x)cos(7x)# using the product rule?

1 Answer
Mar 14, 2018

Answer:

#d/dx f(x) = (1-12e^(4x))/(2sqrt(x-3e^(4x)))(cos(7x))+(-7sin(7x))(sqrt(x-3e^(4x)))#

Explanation:

I'm assuming your original post meant #sqrt(x-3e^(4x))cos(7x)# and this was a simple typo in inferring this question. Therefore let us begin by recalling the product rule.

#(fg)'=f'g +fg'# where #f and g# are differentiable.

Let us define #f and g#

#f(x) = sqrt(x-3e^(4x))#
#g(x)=cos(7x)#

To simplify this problem into bite size parts let us differentiate each function separately and plug in our response at the end of the problem.

Therefore,
#d/dx f(x)# requires us to use the chain rule.

where our inner function is #x^(1/2)# and our outer function is #(x-3e^(4x))#

recall #(f circ g)' = (f' circ g)* g'#

#f' = (1/2)(x^(-1/2))# or
#(1)/(2sqrt(x))# of our inner function where we must plug in #g = (x-3e^(4x))#
#(1)/(2sqrt((x-3e^(4x)))# now find #g'#

#g = (x-3e^(4x))#

#g' = 1-12e^(4x)# recalling that differentiation splits across sums and differences making two derivatives. Noting #d/dx e^x = (x')e^x#

Where our #x=(4x)#

Furthermore,
#d/dx f(x) = (1-12e^(4x))/(2sqrt(x-3e^(4x))#

Now let us find #d/dx g(x) = cos(7x)# recalling that we must use the chain rule again.

#d/dx g(x) = -7sin(7x)# where

#f = cos(x) and g=7x#
#(f circ g)' = (f' circ g)* g'#

#d/dx cos(x) = -sin(x)#

#d/dx 7x = 7#

Recall our original problem #sqrt(x-3e^(4x))cos(7x)#
#(fg)'=f'g +fg'# where #f and g# are differentiable.

#f = sqrt(x-3e^(4x)) and g=cos(7x)# also

#f' = (1-12e^(4x))/(2sqrt(x-3e^(4x))# and #g' =-7sin(7x)#

#(1-12e^(4x))/(2sqrt(x-3e^(4x)))(cos(7x))+(-7sin(7x))(sqrt(x-3e^(4x)))#