1. ## Implicit Differentiation help

so I am trying to find $\displaystyle \frac {dy}{dx}$ for $\displaystyle x=tan(y)$
here's some of my work
not sure if I'm right but I'm applying the product rule to find it

$\displaystyle u=tan$

$\displaystyle u'=sec^2$

$\displaystyle v=y$

$\displaystyle v'= \frac {dy}{dx}$

$\displaystyle tan \frac {dy}{dx} + sec^2 y=1$

the 1 is from the derivative of x

then I subtracted $\displaystyle sec^2 y$ from both sides

which is where I got stuck

the answer for the problem is supposed to be $\displaystyle cos^2 y$ but I don't know how to get there...

2. Well, it looks like you did some extra work. The derivative of tan(y) = sec^2(y) dy. Thus we have:

x = tan(y)

dx = sec^2(y) dy

cos^2(y) = dy/dx

We get to the final answer by dividing both sides by dx and then dividing both sides by sec^2(y).

I hope this helps.

3. after doing what you said, I get $\displaystyle \frac {1}{sec^2 x} = \frac {dy}{dx}$
I don't get how that is $\displaystyle cos^2 y$
is that a trig identity?

4. sec^2(y) is the same as saying 1/cos^2(y). So, having 1/sec^2(y) is the same as:

1
---
1
----
cos^2(y)

We multiply by:

cos^2(y)
----------
cos^2(y)

and we get cos^2(y).

I hope this clears things up!