Question #056a7

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Answer 1 · 2017-11-05T18:04:33

$sec^2(x)$

Rewrite the expression: $1+tanx=1+sinx/cosx$

= $d/dx (1+sinx/cosx)$

differentiate the sum term by term

= $d/dx(1)+d/dx(sinx/cosx)$

the derivative of 1 is zero

$d/dx(sinx/cosx)$

Use the quitient rule

$(d/dx((sinx))cosx-(d/dx(cosx)sinx))/(cos^(x))$

simplify

$(cos^2(x)+sin^2(x))/(cos^2(x)$

use the Pythagorean identity $cos^2(x)+sin^2(x)=1$

$=(1/cos^2(x))$

$=sec^2(x)$