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

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

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

yes

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

\[2\tan (\theta)\sec (\theta)\]

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

\[4e^\theta\]

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

\[\sec (\theta)^2\]

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

what about the tan\[\theta\]

The one in the middle?

yes

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.

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

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

yay team \c:/

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

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.

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