Question #95e51

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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

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