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Calculus1 11 Online
OpenStudy (anonymous):

Whats the derivative of tan^theta - theta?

OpenStudy (anonymous):

\[y=\tan^\theta-\theta\] Is this what you mean?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

So we are taking tangent of -theta to the power of theta.

OpenStudy (anonymous):

Im supposed to differentiate \[g(\theta)=e^{\theta}(\tan \theta- \theta)\]

OpenStudy (anonymous):

\[g'(\theta)=[e^\theta*(\tan(\theta)-\theta)]+[e^\theta*\sec^2(\theta)]\rightarrow g'(\theta)=e^\theta[(\tan(\theta)-\theta)+(\sec^2(\theta)]\] @meaghan716

OpenStudy (anonymous):

Btw, I'm assuming that theta is independent and not a function of another variable.

OpenStudy (noelgreco):

In the second set of brackets on the left you need \[[e ^{\theta}*(\sec ^{2}\theta - 1)]\]

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