Question

How do you find the first and second derivatives of `f(x)=-2/(x^2-2x)+(x+2)/(x-3)` using the quotient rule?

Answer 1

`f'(x)=(4x-4)/(x^2-2x)^2-5/(x-3)^2`
`f''(x)=(4(x^2-2x)-4(2x-2)^2)/(x^2-2x)^3+10/(x-3)^3`

Explanation:

`f(x)=-2/(x^2-2x)+(x+2)/(x-3)`

`f'(x)=d/dx[-2/(x^2-2x)+(x+2)/(x-3)]`

`color(white)(f'(x))=d/dx[-2/(x^2-2x)]+d/dx[(x+2)/(x-3)]`

`color(white)(f'(x))=-(d/dx[2] (x^2-2x)-2d/dx[x^2-2x])/(x^2-2x)^2+((x-3)d/dx[x+2]-(x+2)d/dx[x-3])/(x-3)^2`

`color(white)(f'(x))=-(0(x^2-2x)-2(2x-2))/(x^2-2x)^2+(1(x-3)-1(x+2))/(x-3)^2`

`color(white)(f'(x))=-(-2(2x-2))/(x^2-2x)^2+(x-3-x-2)/(x-3)^2`

`color(white)(f'(x))=(4x-4)/(x^2-2x)^2+(-5)/(x-3)^2`

`color(white)(f'(x))=(4x-4)/(x^2-2x)^2-5/(x-3)^2`

`f''(x)=d/dxf'(x)=d/dx[(4x-4)/(x^2-2x)^2-5/(x-3)^2]`

`color(white)(f''(x))=d/dx[(4x-4)/(x^2-2x)^2]-d/dx[5/(x-3)^2]`

`color(white)(f''(x))=(d/dx[4x-4] (x^2-2x)^2-(4x-4)d/dx[(x^2-2x)^2])/((x^2-2x)^2)^2-(d/dx[5] (x-3)^2-5d/dx[(x-3)^2])/((x-3)^2)^2`

`color(white)(f''(x))=(4(x^2-2x)^2-(4x-4)d/dx[(x^2-2x)^2])/(x^2-2x)^4-(0(x-3)^2-5d/dx[(x-3)^2])/(x-3)^4`

`color(white)(f''(x))=(4(x^2-2x)^2-(4x-4)(2*(x^2-2x)^(2-1)d/dx[x^2-2x]))/(x^2-2x)^4-(0(x-3)^2-5(2*(x-3)^(2-1)d/dx[x-3]))/(x-3)^4`

`color(white)(f''(x))=(4(x^2-2x)^2-(4x-4)(2(x^2-2x)(2x-2)))/(x^2-2x)^4-(-5(2(x-3)(1)))/(x-3)^4`

`color(white)(f''(x))=(4(x^2-2x)^2-2(4x-4)(x^2-2x)(2x-2))/(x^2-2x)^4+(10(x-3))/(x-3)^4`

`color(white)(f''(x))=(4(x^2-2x)^2-4(2x-2)(x^2-2x)(2x-2))/(x^2-2x)^4+10/(x-3)^3`

`color(white)(f''(x))=(4(x^2-2x)^2-4(2x-2)^2(x^2-2x))/(x^2-2x)^4+10/(x-3)^3`

`color(white)(f''(x))=(4(x^2-2x)-4(2x-2)^2)/(x^2-2x)^3+10/(x-3)^3`