Question #95e51

1 Answer
Nam D.
Jan 8, 2018

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

Explanation:

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))#

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