Question

How do you find `(d^2y)/(dx^2)` given `y=(2x+5)^(-1/2)`?

Answer 1

`(d^2y)/(dx^2)=3/(2x+5)^(5/2)`

Explanation:

To find the `color(blue)"first derivative" dy/dx` differentiate using the `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))to(A)`

let `u=2x+5rArr(du)/(dx)=2`

and `y=u^(-1/2)rArr(dy)/(du)=-1/2u^(-3/2)`

substitute into (A) changing u back to x.

`rArrdy/dx=-1/2u^(-3/2)xx2=-(2x+5)^(-3/2)`

To find the `color(blue)"second derivative "(d^2y)/(dx^2)` repeat the process above on `dy/dx` using the `color(blue)"chain rule"`

`u=2x+5rArr(du)/(dx)=2`

`y=-u^(-3/2)rArr(dy)/(du)=3/2u^(-5/2)`

`rArr(d^2y)/(dx^2)=3(2x+5)^(-5/2)=3/(2x+5)^(5/2)`