What are the first and second derivatives of # g(x) = lncscx#?

Question

1195 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-02T17:19:25

#g'(x)=-cotx,g''(x)=csc^2x#

According to the chain rule,

#d/dx(ln(u))=1/u*(du)/dx#

Thus,

#d/dx(ln(cscx))=1/cscx*d/dx(cscx)#

#=>1/cscx*(-cscxcotx)#

#=>-cotx=g'(x)#

To find the second derivative, know that #d/dx(cotx)=-csc^2x#.

Thus,

#d/dx(-cotx)=csc^2x=g''(x)#