Question

How do you find the derivative of `ln[sqrt(2+x^2)/(2-x^2)]`?

Answer 1

see below

Explanation:

Use the following Properties of Logarithms to expand the problem first before finding the derivative

1. `color(red)(log_b(xy)=log_bx+log_by`
2. `color(red)(log_b(x/y)=log_bx-log_by`
3. `color(red)(log_b x^n =n log_bx`

`y=ln[(sqrt(2+x^2))/(2-x^2)]=ln[((2+x^2)^(1/2))/(2-x^2)]`

`=ln(2+x^2)^(1/2)-ln(2-x^2)`

`=1/2 ln(2+x^2)-ln(2-x^2)`

`color(blue)(y'=1/2 *(2x)/(2+x^2)-(-2x)/(2-x^2)`

`color(blue)(y'=1/cancel 2 *(cancel 2x)/(2+x^2)-(-2x)/(2-x^2)`

`color(blue)(y'=x/(2+x^2)+(2x)/(2-x^2)`

`color(blue)(y'=(x(2-x^2)+2x(2+x^2))/((2+x^2)(2-x^2))`

`color(blue)(y'=(2x-x^3+4x+2x^3)/((2+x^2)(2-x^2))`

`color(blue)(y'=(x^3+6x)/((2+x^2)(2-x^2))`