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

3 Answers
Apr 23, 2018

Answer:

#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

Answer:

#=>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

Answer:

#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)#