Mathematics
OpenStudy (anonymous):

Differentiate the following function. h(θ) = 2 sec θ + 4eθ tan θ

zepdrix (zepdrix):

$\Large h(\theta)=2\sec \theta+4e^{\theta}\tan \theta$ $\Large h'(\theta)=\color{royalblue}{(2\sec \theta)'}+\color{royalblue}{(4e^{\theta})'}\tan \theta+4e^{\theta}\color{royalblue}{(\tan \theta)'}$

zepdrix (zepdrix):

Looks like we need to apply the product rule, there is our setup. We need to take the derivative of the blue parts. Understand what's going on so far? :o

OpenStudy (anonymous):

yes

zepdrix (zepdrix):

Do you know any of those derivatives? :) secant? :D tangent? :D e^x? :x

OpenStudy (anonymous):

$2\tan (\theta)\sec (\theta)$

zepdrix (zepdrix):

$\Large h'(\theta)=\color{orangered}{(2\sec \theta \tan \theta)}+\color{royalblue}{(4e^{\theta})'}\tan \theta+4e^{\theta}\color{royalblue}{(\tan \theta)'}$Looks good so far :)

OpenStudy (anonymous):

$4e^\theta$

zepdrix (zepdrix):

$\Large h'(\theta)=\color{orangered}{(2\sec \theta \tan \theta)}+\color{orangered}{(4e^{\theta})}\tan \theta+4e^{\theta}\color{royalblue}{(\tan \theta)'}$Good good good.

OpenStudy (anonymous):

$\sec (\theta)^2$

zepdrix (zepdrix):

$\Large h'(\theta)=\color{orangered}{(2\sec \theta \tan \theta)}+\color{orangered}{(4e^{\theta})}\tan \theta+4e^{\theta}\color{orangered}{(\sec^2\theta)}$Yay ben!

OpenStudy (anonymous):

what about the tan$\theta$

zepdrix (zepdrix):

The one in the middle?

OpenStudy (anonymous):

yes

zepdrix (zepdrix):

Product rule tells us to do this:$\Large (fg)' \quad=\quad f'g+fg'$ We leave the first g alone, and the second f alone.

zepdrix (zepdrix):

You can think of the tangent in the middle as your first g.

OpenStudy (anonymous):

oh yeah sorry forgot you did that for me in the beginning

zepdrix (zepdrix):

yay team \c:/

OpenStudy (anonymous):

i typed this into webassign but it says im wrong did i type it wrong? (2secθtanθ)+(4eθ)tanθ+4eθ(sec2θ)

zepdrix (zepdrix):

Hmm did you do your exponents correctly? (2secθtanθ)+(4e^θ)tanθ+4e^θ(secθ)^2 Maybe they're just not showing up on this page.