Question

How do you differentiate `h(x)=((x^3+1)sqrtx)/x^2` using the quotient rule?

Answer 1

`h'(x) = (3sqrtx(x^3-1))/(2x^3)`

Explanation:

`h(x) = ((x^3+1)sqrtx)/x^2`

`= (x^3+1)/x^(3/2)`

Apply the quotient rule.

`h'(x) = (x^(3/2)* (3x^2) - (x^3 +1) * (3/2x^(1/2)))/x^3`

`= (3x^(7/2) - 3/2sqrtx(x^3+1))/x^3`

`= (6x^(7/2) - 3sqrtx(x^3+1))/(2x^3)`

`= (6x^(7/2)-3x^(7/2) -3sqrtx)/(2x^3)`

`= (3x^(7/2) - 3sqrtx)/(2x^3)`

`= (3sqrtx(x^3-1))/(2x^3)`

Answer 2

`h^'(x)=(3(x^3-1))/((2sqrtx)x^2)=3/2*x^(1/2)-3/2*x^(-5/2)`, (simplify)
OR
`h(x)=((x^3+1)sqrt(x))/x^2=(x^3sqrtx)/x^2+(sqrtx)/x^2=x^(3+1/2-2)+x^(1/2-2)=x^(3/2)+x^(-3/2)=>h^'(x)=3/2*x^(1/2)-3/2*x^(-5/2)`

Explanation:

Here, `h(x)=((x^3+1)sqrt(x))/x^2`
Let,`f(x)=(x^3+1)sqrtxandg(x)=x^2=>g^'(x)=2x`
`andf^'(x)=(x^3+1)d/(dx)(sqrtx)+sqrtx*d/(dx)((x^3+1)`
`=>f^'(x)=(x^3+1)*1/(2sqrtx)+sqrtx*3x^2=(x^3+1+2x*3x^2)/(2sqrtx)`
`=>f^'(x)=(x^3+1+6x^3)/(2sqrtx)=(7x^3+1)/(2sqrtx)`
We know that, If `h(x)=(f(x))/(g(x)),then,`
`h^'(x)=(g(x)*f^'(x)-f(x)*g^'(x))/([g(x)]^2)=(x^2((7x^3+1)/(2sqrtx))-(x^3+1)sqrt(x)*2x)/((x^2)^2)=(x^2(7x^3+1)-(x^3+1)2x*2x)/((2sqrtx)*x^4)`
`=(7x^5+x^2-(x^3+1)4x^2)/((2sqrtx)x^4)=(7x^5+x^2-4x^5-4x^2)/((2sqrtx)x^4)=(3x^5-3x^2)/((2sqrtx)x^4)=(3x^2(x^3-1))/((2sqrtx)x^4)=(3(x^3-1))/((2sqrtx)x^2)`