Derivative of Trig Function using quotient rule Help

• Sep 25th 2012, 07:17 AM
psilver1
Derivative of Trig Function using quotient rule Help
Hey guys need some help here. The trig identities are killing me so far this semester. i have to use the quotient rule. Here is the problem and where i have gotten to.

y= sinx+cosx/cosx
y'= f'g-fg'/g^2
= {[(cosx-sinx)(cosx)]-[(sinx+cosx)(-sinx)]}/cos^2x
= (cos^2x-sinxcosx)-(-sin^2x-sinxcosx)/cos^2x
=(cos^2x-sinxcosx+sin^2x+sinxcosx/cos^2x
=cos^2x+sin^2x/cos^2x
= (cos^2x/cosx)+(sin^2x/cosx)
= am i even close. thanks for the help
• Sep 25th 2012, 07:39 AM
Siron
Re: Derivative of Trig Function using quotient rule Help
Quote:

Originally Posted by psilver1
y= sinx+cosx/cosx
y'= f'g-fg'/g^2
= {[(cosx-sinx)(cosx)]-[(sinx+cosx)(-sinx)]}/cos^2x
= (cos^2x-sinxcosx)-(-sin^2x-sinxcosx)/cos^2x
=(cos^2x-sinxcosx+sin^2x+sinxcosx/cos^2x
=cos^2x+sin^2x/cos^2x

This is correct. To simplify the expression use the identity $\sin^2(x)+\cos^2(x)=1$.

Note that
$y = \frac{\sin(x)+\cos(x)}{\cos(x)} = \frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\cos(x)} = \tan(x)+1$
and $\frac{d}{dx}y = \frac{d}{dx}(\tan x+1) = \frac{1}{\cos^2(x)}$