Question

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

Answer 1

`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)`