How do you differentiate #g(x) = sqrt(x^3-4)cos2x# using the product rule?

Question

Isabella Ackerman · Accepted Answer

#g'(x) = (3x^2 * cos 2x) /(2 sqrt(x^3 - 4)) - 2 sin (2x) * sqrt(x^3-4) #

For #g(x) = u(x) v(x)#, the product rule states that the derivative can be computed as follows:

# g'(x) = u'(x) v(x) + u(x) v'(x)#

As for you,

#u(x) = sqrt(x^3-4)" "# and

#v(x) = cos 2x#

So, first thing to do is compute the derivatives of #u(x)# and #v(x)# using the chain rule:

#u(x) = sqrt(x^3-4) = (x^3-4)^(1/2)# # color(white)(xxxxxxxxxxx) => u'(x) = 1/2 (x^3 - 4)^(- 1 /2 ) * 3x^2 = 1/(2 sqrt(x^3 - 4)) * 3x^2 #

#v(x) = cos 2x color(white)(xx) => v'(x) = - sin (2x) * 2#

Applying the formula is all that is left to do now!

#g'(x) = u'(x) * v(x) + u(x) * v'(x)#

# color(white)(xxxx) = 1/(2 sqrt(x^3 - 4)) * 3x^2 * cos 2x - 2 sin (2x) * sqrt(x^3-4) #

# color(white)(xxxx) = (3x^2 cos 2x) /(2 sqrt(x^3 - 4)) - 2 sin (2x) * sqrt(x^3-4) #

Emily Ammerman · Answer

undefined To differentiate $ g(x) = \sqrt{x^3 - 4} \cos(2x) $ using the product rule, follow these steps:

1. Identify the two functions within the product: $ f(x) = \sqrt{x^3 - 4} $ and $ h(x) = \cos(2x) $.
2. Apply the product rule: $ g'(x) = f(x)h'(x) + f'(x)h(x) $.
3. Find the derivatives of $ f(x) $ and $ h(x) $:
   - $ f'(x) = \frac{1}{2\sqrt{x^3 - 4}} \cdot \frac{d}{dx}(x^3 - 4) $.
   - $ h'(x) = -2\sin(2x) $.
4. Plug the derivatives and original functions into the product rule formula and simplify to get $ g'(x) $.