Let's find the first derivative.
Using quotient rule
\(f(x)=\frac{g(x)}{h(x)}\)
\(f'(x)=\frac{g'(x)*h(x)-g(x)*h'(x)}{h^2(x)}\)
then:
\(y=\frac{x+2}{x-3}\)
\(f(x)=y\)
\(g(x)=x+2\)
\(g'(x)=1\)
\(h(x)=x-3\)
\(h'(x)=1\)
Let's replace
\(f'(x)=\frac{g'(x)*h(x)-g(x)*h'(x)}{h^2(x)}\)
\(y'=\frac{1*(x-3)-(x+2)*1}{(x-3)^2}\)
\(y'=\frac{x-3-x-2}{(x-3)^2}\)
\(\boxed{y'=\frac{-5}{(x-3)^2}}\)
the second derivative is the derivative of our first derivative.
\(y'=\frac{-5}{(x-3)^2}\)
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 \((x-3)\)
\(f'[g(x)]=10*u^{-3}\)
\(f'[g(x)]=10*(x-3)^{-3}\)
\(\boxed{\boxed{y''=\frac{10}{(x-3)^{3}}}}\)
