Question

How do you find the first and second derivative of `lnx^2/x`?

Answer 1

`((lnx^2)/x)''=(2(-3+lnx^2))/x^3`

Explanation:

The derivative of the quotient `(u(x))/(v(x))` is determined using Quotient Rule

`color(blue)(((U(x))/(V(x)))'=(u'(x)v(x)-v'(x)u(x))/v^2`

`color(red)((ln(u(x)))'=((u(x))')/(u(x))`

`((lnx^2)/x)'=color(blue)(((lnx^2)'*x-x'(lnx^2))/x^2)`

`((lnx^2)/x)'=(color(red)((2x)/x^2)*x-1(lnx^2))/x^2`

`((lnx^2)/x)'=((2x^2)/x^2-1(lnx^2))/x^2`
`color(purple)(((lnx^2)/x)'=(2-lnx^2)/x^2)`

`((lnx^2)/x) '' =((color(purple)(lnx^2)/x)')'`

`((lnx^2)/x)''=((color(purple)((lnx^2)/x)')'`
`((lnx^2)/x)''=((2-lnx^2)/x^2))'`

`((lnx^2)/x)''=color(blue)(((2-lnx^2)'x^2-(x^2)'(2-lnx^2))/x^4 )`
`((lnx^2)/x)''=((0-(2x)/x^2)x^2-(2x)(2-lnx^2))/x^4`
`((lnx^2)/x)''=(((-2x)/cancelx^2)cancelx^2-(2x)(2-lnx^2))/x^4`
`((lnx^2)/x)''=(-2x-(2x)(2-lnx^2))/x^4`
`((lnx^2)/x)''=(-2x-4x+2xlnx^2)/x^4`
`((lnx^2)/x)''=(-6x+2xlnx^2)/x^4`
`((lnx^2)/x)''=2x(-3+lnx^2)/x^4`
`((lnx^2)/x)''=(2(-3+lnx^2))/x^3`