How do you find the derivative of $f (x) = 4 x + π tan (x)$ $f(x) = 4x + piTan(x)$ ?

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Answer 1 · 2017-12-15T11:52:29

$4+π(1+tan^2x)$

$f'(x)=(4x+πtan(x))' = 4(x)'+π(tanx)' = 4+π(1+tan^2x)$

$(tanx)'=(sinx/cosx)'$ $=((sinx)'cosx-sinx(cosx)')/cos^2x =$

$(cos^2x+sin^2x)/cos^2x = 1/cos^2x = 1+tan^2x$

& $sin^2x+cos^2x=1 <=> sin^2x/cos^2x+1=1/cos^2x <=> 1 + tan^2x = 1/cos^2x$

How do you find the derivative of f(x) = 4x + piTan(x) f(x)=4x+πtan(x)f(x) = 4x + piTan(x) ?