Question

use implicit differentiation to find dy/dx:

Answer 1

\[\cos(r)+\cot(\theta)=e ^{r \theta}\]

Answer 2

Please help! I'm so lost in my calc class and I don't start with my tutor until tomorrow.. :(

Answer 3

You want to find dy/dx or dr/dtheta or dtheta/dr?

Answer 4

dr/dtheta

Answer 5

Okay:
So firstly you want to differentiation both sides:
\[\frac{d}{d \theta}(\cos(r)+\cot(\theta))=\frac{d}{d \theta}e^{r \theta}\]
So for the first term:
\[\frac{d}{d \theta}\cos(r)=-\sin(r) * \frac{d r }{d \theta}\]
For the second term there is no "r" so its a "normal" derivative:
\[\frac{d}{d \theta}\cot(\theta)=-\csc^2(\theta)\]
Now the hardest term is the right side. When you use the chain rule you differentiate the outside, and the inside (via the chain rule). So you would have:
\[\frac{d}{d \theta}e^{r \theta}=e^{r \theta} * \frac{d}{d \theta}(r \theta)=e^{r \theta}(\frac{dr}{d \theta} \theta+r (1))\]
implies:
\[-\sin(r) \frac{dr}{d \theta}-\csc^2(\theta)=e^{r \theta}(\frac{dr}{d \theta} \theta+r)\]
Now you need simply solve for dr/d theta

Answer 6

thank you so much!

Answer 7

No problem!