Let’s find the first derivative.

UsingÂ 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 \( (x-3) \)

\( f’[g(x)]=10*u^{-3} \)

\( f’[g(x)]=10*(x-3)^{-3} \)

\( \boxed{\boxed{y’’=\frac{10}{(x-3)^{3}}}} \)

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

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