Question

How do you differentiate `f(x)=sqrt(ln(x^2+3)` using the chain rule.?

Answer 1

`f'(x)=(x(ln(x^2+3))^(-1/2))/(x^2+3)=x/((x^2+3)(ln(x^2+3))^(1/2))=x/((x^2+3)sqrt(ln(x^2+3)))`

Explanation:

We are given:
`y=(ln(x^2+3))^(1/2)`
`y'=1/2*(ln(x^2+3))^(1/2-1)*d/dx[ln(x^2+3)]`
`y'=(ln(x^2+3))^(-1/2)/2*d/dx[ln(x^2+3)]`

`d/dx[ln(x^2+3)]=(d/dx[x^2+3])/(x^2+3)`

`d/dx[x^2+3]=2x`

`y'=(ln(x^2+3))^(-1/2)/2*(2x)/(x^2+3)=(x(ln(x^2+3))^(-1/2))/(x^2+3)=x/((x^2+3)(ln(x^2+3))^(1/2))=x/((x^2+3)sqrt(ln(x^2+3)))`