How do you differentiate #g(x) =x^3 sqrt(4-x)# using the product rule?
3 Answers
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)#
Explanation:
We are given:
We apply the product rule as follows:
Simplifying:
Explanation:
We can use the product rule with the form;
Where we let one term equal to
Here;
Let
And we can do the same for
Let
Just like
By using chain rule;
Now we can substitute these equations in blue into the equation from the start;