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