Question

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

Answer 1

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

Answer 2

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

Answer 3

yes

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

\[4e^\theta\]

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

what about the tan\[\theta\]

Answer 12

The one in the middle?

Answer 13

yes

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

yay team \c:/

Answer 18

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

Answer 19

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.