Question

How do you find the second derivative of `ln(x^2+5x)` ?

Answer 1

`-(2x^2+10x+25)/(x^2+5x)^2`

Explanation:

First, we need to know the derivative of `ln(x)`.

`d/dxln(x)=1/x`

So, since we have a function embedded within `ln(x)`, we must use the chain rule to differentiate it.

`d/dxln(f(x))=1/f(x)*f'(x)`

Thus:

`d/dxln(x^2+5x)=1/(x^2+5x)*d/dx(x^2+5x)`

`=(2x+5)/(x^2+5x)`

Now, to differentiate this again, use the quotient rule:

`d/dxf(x)/(g(x))=(f'(x)g(x)-f(x)g'(x))/(g(x))^2`

Thus:

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

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

`=(2x^2+10x-(4x^2+20x+25))/(x^2+5x)^2`

`=(-2x^2-10x-25)/(x^2+5x)^2`

Answer 2

`-(2x^2+10x+25)/(x^2+5x)^2`

Explanation:

First derivative: `(1/(x^2+5x))*(2x+5)=(2x+5)/(x^2+5x)`

Second derivftive: `(2(x^2+5x)-(2x+5)(2x+5))/(x^2+5x)^2`

`=-(2x^2+10x+25)/(x^2+5x)^2`

Answer 3

`(-2x^2-10x-25)/(x^2+5x)^2`

Explanation:

To find the first derivative use the `color(blue)"chain rule"`

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

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

then `y=lnurArr(dy)/(du)=1/u`

substitute these values into (A) changing u back to x.

`rArrdy/dx=1/uxx(2x+5)=(2x+5)/(x^2+5x)`
`color(blue)"-----------------------------------------------------"`

To find the second derivative use the `color(blue)"quotient rule"`

If `f(x)=(g(x))/(h(x))" then"`

`color(red)(bar(ul(|color(white)(a/a)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(a/a)|)))`

here `g(x)=2x+5rArrg'(x)=2`

and `h(x)=x^2+5xrArrh'(x)=2x+5`

`f'(x)=((x^2+5x).2-(2x+5)(2x+5))/(x^2+5x)^2`

simplifying the numerator.

`=(2x^2+10x-4x^2-20x-25)/(x^2+5x)^2`

The second derivative is therefore.

`=(-2x^2-10x-25)/(x^2+5x)^2=(-(2x^2+10x+25))/(x^2+5x)^2`