Question

What are the first and second derivatives of `g(x) =ln(sinx)-sin(lnx)`?

Answer 1

`g'(x)=cotx-(cos(lnx))/x`

`g''(x)=csc^2x+(sin(lnx)+cosln(x))/x^2`

Explanation:

According to the chain rule:

`d/dx[ln(u)]=(u')/u`

`d/dx[sin(u)]=u'cos(u)`

Thus,

`d/dx[ln(sinx)]=(d/dx[sinx])/sinx=cosx/sinx=cotx`

`d/dx[sin(lnx)]=d/dx[lnx]cos(lnx)=(cos(lnx))/x`

`g'(x)=cotx-(cos(lnx))/x`

To find `g''(x)`, find the derivative of each part.

`d/dx[cotx]=-csc^2x`

`d/dx[(cos(lnx))/x]=(xd/dx[cos(lnx)]-cos(lnx)d/dx[x])/x^2`

Find each derivative.

`d/dx[cos(lnx)]=-sin(lnx)/x`

`d/dx[x]=1`

Plug back in:

`d/dx[(cos(lnx))/x]=(-sin(lnx)-cosln(x))/x^2`

Thus,

`g''(x)=csc^2x+(sin(lnx)+cosln(x))/x^2`