Question

y=log(secx +tanx) differtiate with respect to sec x where x=π/4

Answer 1

Answer: y'=sec(x)

Full explanation:

Suppose, y=ln(f(x))

Using chain rule, y'=1f(x)⋅f'(x)

Similarly, if we follow for the problem, then

y'=1sec(x)+tan(x)⋅(sec(x)+tan(x))'

y'=1sec(x)+tan(x)⋅(sec(x)tan(x)+sec2(x))

y'=1sec(x)+tan(x)⋅sec(x)(sec(x)+tan(x))

y'=sec(x)