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

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 $$(x-3)$$

$$f’[g(x)]=10*u^{-3}$$

$$f’[g(x)]=10*(x-3)^{-3}$$

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

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