Let's find the first derivative.
Using quotient rule
[tex]f(x)=\frac{g(x)}{h(x)}[/tex]
[tex]f'(x)=\frac{g'(x)*h(x)-g(x)*h'(x)}{h^2(x)}[/tex]
then:
[tex]y=\frac{x+2}{x-3}[/tex]
[tex]f(x)=y[/tex]
[tex]g(x)=x+2[/tex]
[tex]g'(x)=1[/tex]
[tex]h(x)=x-3[/tex]
[tex]h'(x)=1[/tex]
Let's replace
[tex]f'(x)=\frac{g'(x)*h(x)-g(x)*h'(x)}{h^2(x)}[/tex]
[tex]y'=\frac{1*(x-3)-(x+2)*1}{(x-3)^2}[/tex]
[tex]y'=\frac{x-3-x-2}{(x-3)^2}[/tex]
[tex]\boxed{y'=\frac{-5}{(x-3)^2}}[/tex]
the second derivative is the derivative of our first derivative.
[tex]y'=\frac{-5}{(x-3)^2}[/tex]
We can write this function as
[tex]y'=-5*(x-3)^{-2}[/tex]
now we have to use the Chain rule's
[tex]f(u)=-5u^{-2}[/tex]
and
[tex]g(x)=x-3[/tex]
[tex]f[g(x)]=-5*(x-3)^{-2}[/tex]
then
[tex]f'[g(x)]=f'(u)*g'(x)[/tex]
[tex]f'(u)=-5*(-2)*u^{-3}=10*u^{-3}[/tex]
[tex]g'(x)=1[/tex]
[tex]f'[g(x)]=f'(u)*g'(x)[/tex]
Let's replace
[tex]f'[g(x)]=10*u^{-3}*1[/tex]
Let's change u by [tex](x-3)[/tex]
[tex]f'[g(x)]=10*u^{-3}[/tex]
[tex]f'[g(x)]=10*(x-3)^{-3}[/tex]
[tex]\boxed{\boxed{y''=\frac{10}{(x-3)^{3}}}}[/tex]
