# How do you differentiate r=tantheta/(theta^3+theta+1)?

Nov 28, 2017

Depends if you want $\frac{\mathrm{dr}}{d \theta}$ or the slope of the tangent line.

#### Explanation:

Provided that you simply want the derivative of $r$ with respect to the variable $\theta$, rather than something like the derivative of $x \left(\theta\right) \mathmr{and} y \left(\theta\right)$ with respect to $\theta$, you simply differentiate as normal. In this case, we will need to remember the quotient rule, which states that given $f = \frac{g}{h} , f ' = \frac{\left(g ' h\right) \left(g h '\right)}{h} ^ 2$. With $g = \tan \theta , h = {\theta}^{3} + \theta + 1$, that gives us:

$\frac{\mathrm{dr}}{d \theta} = \frac{\left({\theta}^{3} + \theta + 1\right) {\sec}^{2} \left(\theta\right) - \tan \theta \left(3 {\theta}^{2} + 1\right)}{{\theta}^{3} + \theta + 1} ^ 2$

Which simplifies to:

$\frac{\mathrm{dr}}{d \theta} = {\sec}^{2} \frac{\theta}{\left({\theta}^{3} + \theta + 1\right)} - \tan \theta \frac{3 {\theta}^{2} + 1}{{\theta}^{3} + \theta + 1} ^ 2$

Of note, this is not the slope of the tangent line. To find the slope of the tangent line, one must use parametric equations for $x \mathmr{and} y$. Recall that to convert from polar coordinates to cartesian coordinates, we use the following:

$x = r \cos \theta , y = r \sin \theta$.

The slope of the tangent line will still take the form $\frac{\mathrm{dy}}{\mathrm{dx}}$, which is also equal to $\frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}}$

From our formulae and the product rule (f = g * h, f' = g'h + gh'), $\frac{\mathrm{dy}}{d \theta} = \left(\frac{\mathrm{dr}}{d \theta}\right) \sin \theta + r \cos \theta , \mathmr{and} \frac{\mathrm{dx}}{d \theta} = \frac{\mathrm{dr}}{d \theta} \cos \theta - r \sin \theta$

At this point, what we must do is find dy/(d theta) & dx/(d theta), and divide the former by the latter, and we will have an equation for the slope of the tangent line to the curve. Some of the steps are done below, though it is complex enough that there may be some errors.

We already calculated $\frac{\mathrm{dr}}{d \theta}$ above, meaning that we can plug it in:

$\frac{\mathrm{dy}}{d \theta} = \left({\sec}^{2} \frac{\theta}{\left({\theta}^{3} + \theta + 1\right)} - \tan \theta \frac{3 {\theta}^{2} + 1}{{\theta}^{3} + \theta + 1} ^ 2\right) \sin \theta + \tan \frac{\theta}{{\theta}^{3} + \theta + 1} \cos \theta$
$= \left(\sin \frac{\theta}{{\theta}^{3} + \theta + 1}\right) \left({\sec}^{2} \theta + 1 - \tan \theta \cdot \frac{3 {\theta}^{2} + 1}{{\theta}^{3} + \theta + 1}\right)$

$\frac{\mathrm{dx}}{d \theta} = \left({\sec}^{2} \frac{\theta}{\left({\theta}^{3} + \theta + 1\right)} - \tan \theta \frac{3 {\theta}^{2} + 1}{{\theta}^{3} + \theta + 1} ^ 2\right) \cos \theta - \tan \theta \sin \frac{\theta}{{\theta}^{3} + \theta + 1}$

To find $\frac{\mathrm{dy}}{\mathrm{dx}}$, simply perform the division indicated above. This is left to the reader