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

1 Answer
Mar 14, 2018

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)))