# Implicit diferentiation

1. Oct 17, 2009

### hover

1. The problem statement, all variables and given/known data
Find dy/dx of this equation -
$$y*sec(x)=3x*tan(y)$$

2. Relevant equations

-product rule
-derivative of sec(x) with respect to x is sec(x)tan(x) i believe
-derivative of tan(x) is sec^2(x) i believe

3. The attempt at a solution

$$y*sec(x)=3x*tan(y)$$

$$y*sec(x)*tan(x)+sec(x) \frac{dy}{dx}=3x*sec^2(y)\frac{dy}{dx}+3*tan(y)$$

$$sec(x)\frac{dy}{dx}-3x*sec^2(x)\frac{dy}{dx}=3*tan(y) -y*sec(x)*tan(x)$$

$$\frac{dy}{dx}(sec(x)-3x*sec^2(x))= 3*tan(y) -y*sec(x)*tan(x)$$

$$\frac{dy}{dx}= \frac{3*tan(y) -y*sec(x)*tan(x)}{sec(x)-3x*sec^2(y)}$$

That last line is my solution. I do homework online and every time i enter this it says it is wrong. So where am I going wrong?

2. Oct 17, 2009

### slider142

Your last line is correct. Perhaps the problem is with how you entered it. What is the exact string that you entered?

3. Oct 17, 2009

### hover

I entered it the almost the exact same way except for sec^2 . I typed it in as (sec(y))^2. Are you sure I am right?

4. Oct 17, 2009

### slider142

That's the unambiguous way to write it. I have checked the differentiation independently. In your derivation, you switched one of the y's for an x, but you got the last line, so I assume that was a typo. If your parentheses are all correct in your typed version, I would go ahead and show the professor the problem and your solution personally.

5. Oct 17, 2009

### hover

Ok. Thanks for double checking for me then.

6. Oct 19, 2009

### hover

FIGURED IT OUT. The answer I have here is correct. Its a hundred percent correct. dy/dx does equal that value that i posted in my original post. TOO BAD THEY ASKED FOR dx/dy. Its the classic case of not reading the question correctly :P. I'm so used to taking the derivative of dy/dx I just made the assumption that it was dy/dx. Once I found out that it was dx/dy I was able to get the answer right on the first try :D. Sorry about causing you any trouble if I did.