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

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Answer 1 · 2015-12-12T14:45:24

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

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

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,

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

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