Apricity
Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.