# Math Help - d/dtheta

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

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

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

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

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

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