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

Sagot :

D3xt3R
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]
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