Question

How do you find the second derivative of `z=(x-5)/(x+2)`?

Answer 1

`(d^2z)/dx^2 = -14/(x+2)^3`

Explanation:

Calculate the first derivative using the quotient rule:

`(dz)/dx = ((x+2)d/dx(x-5) -(x-5)d/dx(x+2))/(x+2)^2`

`(dz)/dx = (x+2 -x+5)/(x+2)^2 = 7/(x+2)^2`

Then using the chain rule:

`(d^2z)/dx^2 = d/dx (7/(x+2)^2) = -14/(x+2)^3 d/dx (x+2) = -14/(x+2)^3`

Alternatively you can write the function as:

`z= (x-5)/(x+2) = (x+2-7)/(x+2) = 1-7/(x+2) = 1-7(x+2)^-1`

and you can see that:

`(d^nz)/dx^n = 7(-1)^(n+1)(n!)/(x+2)^(n+1)`

Answer 2

`(d^2z)/dx^2=(-14)/(x+2)^3`

Explanation:

One way is to express the function as a product and differentiate using the product rule.

`rArrz=(x-5)/(x+2)=(x-5)(x+2)^-1`

differentiate using the `color(blue)"product rule"`

`"Given "z=g(x)h(x)" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(dz/dx=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))`

`rArrdz/dx=(x-5).-1(x+2)^-2+(x+2)^(-1).1`

`"to obtain "(d^2z)/(dx^2)" differentiate " dz/dx`

differentiate the first term using the `color(blue)"product rule"`

`rArr(d^2z)/(dx^2)=(x-5).2(x+2)^-3-(x+2)^(-2).1-(x+2)^-2`

`=2(x-5)(x+2)^-3-2(x+2)^-2`

`=2(x+2)^-3(x-5-x-2)`

`=-14/(x+2)^3`