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

1 Answer
Jan 12, 2016

Answer:

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

Explanation:

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)#

In your case,

#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#

Now, the only thing left to do is apply the formula!

#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) #