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

3 Answers
Apr 23, 2018

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)

Apr 23, 2018

=>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)))

Apr 23, 2018

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)