Tiny question about implicit differentiation The following differentiated implicitly with respect to $\theta$:
$3x = \tan \, \theta $
The book says 
$3 dx = \sec^2 \theta \, d \theta$
One could start the calculation like this (I think):
$ \frac{d}{d \theta}3x = \frac{d}{d \theta} \tan \, \theta $
From there I'm not really sure about the steps. 
 A: I don't know what you are asking but I will explain how I view this sort of thing.
What is meant by $$3x = \tan(\theta)$$ is that both sides are the same function of $x$ or $\theta$ and $x$ and $\theta$ are related somehow. Writing out the relation explicitly we have $$3x = \tan(\theta(x)).$$ Now define $f(x) = 3x$ and $g(x) = \tan(\theta(x))$, this equation means that $f = g$.
Applying the derivative operator to both sides $f' = g'$ we have $$3 = \theta'(x)\sec(\theta(x))^2$$ (by the chain rule and derivative of tan = sec^2). Now you can write it as a differential $$3 \mathrm{d}x = \sec(\theta(x))^2 \theta'(x) \mathrm{d}x = \sec(\theta)^2 \mathrm{d}\theta$$ since $\theta'(x) \mathrm{d}x = \mathrm{d}\theta$.
A: This is differential notation.  If $y=f(x)$, then it is customary to write $dy=f'(x)dx$.  What this says is that a small change in $x$, $dx$, produces an approximate change on $y$ of $f'(x)dx$.  In your instance you have $x=f(\theta)$ and so $dx=f'(\theta)d\theta$.  The concept comes into play when you talk about linear approximation and you are using the differentials to approximate error.  So you have 
$$
x=\frac{1}{3}\tan\theta
$$
which gives
$$
dx=\frac{1}{3}\sec^2(\theta) d\theta.
$$
A: It is simply that $$\frac{d}{d \theta} \left( 3x \right) = \frac{d}{d \theta} \left( \tan \theta \right)$$
Then we know that $$\frac{d}{d \theta } \left( \tan \theta \right) = \mbox{sec}^2 \theta $$
and
$$\frac{d}{d \theta} \left( 3x \right) = 3 \frac{dx}{d \theta}$$
(Can you see why?)
