Question

How do you find the derivative of `(x^8(x-23)^(1/2))/(27x^6(4x-6)^8)`?

Answer 1

`=(x^8(432x^5(4x-6)^8(x-23)^(1/2)+(x-23)^(-1/2)-2x^4(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7))))/(2(27x^6(4x-8)^8)^2`

Explanation:

`y=(x^8(x-23)^(1/2))/(27x^6(4x-6)^8)=f(x)/g(x)`

`dy/dx=(f'(x)g(x)-f(x)g'(x))/(f(x)^2)`

`f(x)=x^8(x-23)^(1/2)=h(x)j(x)`
`f'(x)=h'(x)j(x)+h(x)j'(x)`
`h(x)=x^8`
`h'(x)=8x^7`
`j(x)=(x-23)^(1/2)`
`j'(x)=(x-23)^(-1/2)/2`
`f'(x)=8x^7(x-23)^(1/2)+(x^8(x-23)^(-1/2))/2=(16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2=(16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2`

`g(x)=27x^6(4x-6)^8=a(x)s(x)`
`g'(x)=a'(x)s(x)+a(x)s'(x)`
`a(x)=27x^6`
`a'(x)=162x^5`
`s(x)=(4x-6)^8`
`s'(x)=4*8*(4x-6)^7=32(4x-6)^7`
`g'(x)=162x^5(4x-6)^8+27x^6 32(4x-6)^7=162x^5(4x-6)^8+864x^6(4x-6)^7=54x^5(3(4x-6)^8+16x(4x-6)^7)`

`dy/dx=(27x^6(4x-6)^8((16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2)-x^8(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7)))/(27x^6(4x-6)^8)^2`
`=(((27x^6(4x-6)^8 16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2)-x^8(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7)))/(27x^6(4x-6)^8)^2`
`=(27x^6(4x-6)^8 16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2)-2x^8(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7)))/(2(27x^6(4x-6)^8)^2)`
`=(x^8(432x^5(4x-6)^8(x-23)^(1/2)+(x-23)^(-1/2)-2x^4(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7))))/(2(27x^6(4x-8)^8)^2`