What is the derivative of $y = ln (sec (x) + tan (x))$ $y=ln(sec(x)+tan(x))$ ?

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Answer 1 · 2014-07-26T18:18:26

Answer: $y'=sec(x)$

Full explanation:

Suppose, $y=ln(f(x))$

Using chain rule, $y'=1/f(x)*f'(x)$

Similarly, if we follow for the problem, then

$y'=1/(sec(x)+tan(x))*(sec(x)+tan(x))'$

$y'=1/(sec(x)+tan(x))*(sec(x)tan(x)+sec^2(x))$

$y'=1/(sec(x)+tan(x))*sec(x)(sec(x)+tan(x))$

$y'=sec(x)$

Answer 2 · 2015-04-18T00:34:48

Will give you a personal video explanation of how it's done...

Alternatively, you can use these workings...

$ln(secx+tanx)=y$

$e^y=secx+tanx$

$e^y*(dy)/(dx)=secxtanx+sec^2x$

$e^y*(dy)/(dx)=secx(secx+tanx)$

$(dy)/(dx)=(secx(secx+tanx))/e^y$

$(dy)/(dx)=(secx(secx+tanx))/((secx+tanx))$

$(dy)/(dx)=secx$

What is the derivative of y=ln(sec(x)+tan(x))y=ln(sec(x)+tan(x))y=ln(sec(x)+tan(x))?