Question

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

Answer 1

`g'(x)=3x^2sqrt(4-x)-(x^3)/(2sqrt(4-x))`

Explanation:

`"Given "y=f(x)h(x)" then"`

`dy/dx=f(x)h'(x)+h(x)f'(x)larrcolor(blue)"product rule"`

`f(x)=x^3rArrf'(x)=3x^2`

`h(x)=sqrt(4-x)=(4-x)^(1/2)`

`"differentiate using the "color(blue)"chain rule"`

`rArrh'(x)=1/2(4-x)^(-1/2)xxd/dx(4-x)`

`color(white)(rArrh'(x))=-1/(2sqrt(4-x))`

`rArrg'(x)=-(x^3)/(2sqrt(4-x))+3x^2sqrt(4-x)`

Answer 2

`=>g'(x) = ((24-7x)x^2)/(2sqrt(4-x))`

Explanation:

We are given:

`g(x) = x^3sqrt(4-x)`

We apply the product rule as follows:

`g'(x) = d/(dx)[x^3] sqrt(4-x) + x^3d/(dx)[sqrt(4-x)]`

Simplifying:

`g'(x) = [3x^2]sqrt(4-x) + x^3[-1/2(1)/(sqrt(4-x))]`

`g'(x) = 3x^2sqrt(4-x) -(x^3)/(2sqrt(4-x))`

`g'(x) = 3x^2sqrt(4-x)*(2sqrt(4-x))/(2sqrt(4-x))-(x^3)/(2sqrt(4-x))`

`g'(x) = (6x^2(4-x))/(2sqrt(4-x)) -(x^3)/(2sqrt(4-x))`

`g'(x) = (24x^2-6x^3-x^3)/(2sqrt(4-x))`

`g'(x) = (24x^2-7x^3)/(2sqrt(4-x))`

`=>color(blue)(g'(x) = ((24-7x)x^2)/(2sqrt(4-x)))`

Answer 3

`g'(x) = 3x^2sqrt(4-x) - x^3/(2sqrt(4-x)`

Explanation:

We can use the product rule with the form;

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

Where we let one term equal to `u` and the other equal to `v`.
Here;

Let `color(blue)(u = x^3`

`u` would have to be differentiated by multiplying the term by the initial power and decreasing the power by one;

`u' = 3 * x^(3-1)`
`u' = 3x^2`

And we can do the same for `v`;

Let `color(blue)(v = sqrt(4-x)`

Just like `u`, `v` would also have to be differentiated. We can start by changing the square root to the power of `1/2` to make it easier;

`v = (4-x)^(1/2)`

By using chain rule;

`v' = 1/2 * (4-x)^(1/2 - 1)* (4-x)'`

`v' = -1/2 * (4-x)^(-1/2)`

`v' = -1/2 * 1/sqrt(4-x)`

`color(blue)(v' = -1/(2sqrt(4-x))`

Now we can substitute these equations in blue into the equation from the start;

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

`g'(x) = 3x^2*sqrt(4-x) -1/(2sqrt(4-x)) * x^3`

`g'(x) = 3x^2sqrt(4-x) - x^3/(2sqrt(4-x)`