1. ## d/dtheta

h(theta)= 9sec(theta)+9e^(theta)cot(theta)

h'(theta)=d/dtheta(9sec(theta)) + d/dtheta(9e^(theta)cot(theta))

?= 9sec(theta) 9 tan(theta)+ ..... ?

2. Originally Posted by vjelmy
h(theta)= 9sec(theta)+9e^(theta)cot(theta)

h'(theta)=d/dtheta(9sec(theta)) + d/dtheta(9e^(theta)cot(theta))

?= 9sec(theta) 9 tan(theta)+ ..... ?
Since $\displaystyle \theta$ is a constant, we are not going to explicitly differentiate it, since $\displaystyle dk/dx$, where k is a constant, is 0. For the first term $\displaystyle 9sec(\theta)$, remember that $\displaystyle d/dx * sec(x) = sec(x)tan(x)$.

For the second term, remember that $\displaystyle d/dx e^x = dx e^x$. So if $\displaystyle x = \theta cot(\theta)$, then $\displaystyle dx = \theta * -csc^2(\theta)$.

3. Originally Posted by vjelmy
h(theta)= 9sec(theta)+9e^(theta)cot(theta)

h'(theta)=d/dtheta(9sec(theta)) + d/dtheta(9e^(theta)cot(theta))

?= 9sec(theta) 9 tan(theta)+ ..... ?
$\displaystyle \frac{d}{d\theta}9e^{\theta}cot{\theta} = 9[cot{\theta}\frac{d}{d\theta}e^{\theta} + e^{\theta}\frac{d}{d\theta}cot{\theta}]$

$\displaystyle = 9 [ e^{\theta}cot{\theta} - e^{\theta} csc^2{\theta} ]$

4. Originally Posted by macosxnerd101
Since $\displaystyle \theta$ is a constant, we are not going to explicitly differentiate it, since $\displaystyle dk/dx$, where k is a constant, is 0. For the first term $\displaystyle 9sec(\theta)$, remember that $\displaystyle d/dx * sec(x) = sec(x)tan(x)$.

For the second term, remember that $\displaystyle d/dx e^x = dx e^x$. So if $\displaystyle x = \theta cot(\theta)$, then $\displaystyle dx = \theta * -csc^2(\theta)$.
Where did you get the idea that $\displaystyle \theta$ was a constant? $\displaystyle \theta$ is the only variable in the problem! And what in the world does the derivative of $\displaystyle e^x$ have to do with this?