Question #95e51

Question

790 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-08T12:36:52

#f'(x)=(sec(x)tan(x))/(2+sec(x))#

Using the chain rule,

#(df(u))/dx=(df)/(du)*(du)/(dx)#

#f=ln(u)#

#(df)/(du)=1/u#

#u=2+sec(x)#

#(du)/dx=sec(x)tan(x)#

#:.(df(u))/dx=1/u*sec(x)tan(x)=(sec(x)tan(x))/u#

Substitute #u=2+sec(x)# back, we get

#(df(u))/dx=(sec(x)tan(x))/(2+sec(x))#

or

#f'(x)=(sec(x)tan(x))/(2+sec(x))#