Question

Find the second derivative for y=(x+2)/(x-3)

Answer 1

Let's find the first derivative.

Using quotient rule


`f(x)=(g(x))/(h(x))`


`f'(x)=(g'(x)*h(x)-g(x)*h'(x))/(h^2(x))`

then:


`y=(x+2)/(x-3)`

`f(x)=y`


`g(x)=x+2`


`g'(x)=1`


`h(x)=x-3`


`h'(x)=1`

Let's replace


`f'(x)=(g'(x)*h(x)-g(x)*h'(x))/(h^2(x))`


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


`y'=(x-3-x-2)/((x-3)^2)`


`\boxed{y'=(-5)/((x-3)^2)}`

the second derivative is the derivative of our first derivative.


`y'=(-5)/((x-3)^2)`

We can write this function as


`y'=-5*(x-3)^(-2)`

now we have to use the Chain rule's


`f(u)=-5u^(-2)`

and


`g(x)=x-3`


`f[g(x)]=-5*(x-3)^(-2)`

then


`f'[g(x)]=f'(u)*g'(x)`


`f'(u)=-5*(-2)*u^(-3)=10*u^(-3)`


`g'(x)=1`


`f'[g(x)]=f'(u)*g'(x)`

Let's replace


`f'[g(x)]=10*u^(-3)*1`

Let's change u by
`(x-3)`


`f'[g(x)]=10*u^(-3)`


`f'[g(x)]=10*(x-3)^(-3)`


`\boxed{\boxed{y''=(10)/((x-3)^(3))}}`