## anonymous one year ago Find dy/dx of x cos(y) = 1 at the point (-1,pi) using implicit differentiation

1. anonymous

I solved this problem on a test and got it right, but now I don't remember how to do it. Find $\frac{ dy }{ dx } x \cos(y)$ at the point (−1,π)

2. dessyj1

It seems like we will have to use the product rule. Are you familiar with the product rule?

3. nincompoop

do you want to just get the general derivative implicitly first then add the values later

4. anonymous

Yes. But do I keep the x as it is or do I rplacee it with -1?

5. nincompoop

product rule, kid

6. anonymous

@nincompoop I know the product rule

7. dessyj1

Do you need help with implicitly deriving the left side of your equation?

8. anonymous

I know ow to do the implicit part. This is what I did on the test (and got it correct but now it doesn't make any sense) Step 1: (-1) (-sin(y))y' + cos(y) (-1) = 0 Step 2: sin(y) y' - cos(y) Step 3: sin(y) y' = cos(y) Step 4: $y' = \frac{ \cos(y) }{ \sin(y) }$ In step 1 shouldn't it be + cos(y) (0) ? but I didn't write that and still got the answer correct

9. dessyj1

It would not be cos(y)(0) because the derivative of x is 1

10. dessyj1

Also, of you were following the f'g +fg' you will realize that you forgot to multiply y'sin(y) by an x.

11. dessyj1

Remember that the derivative of a constant is 0

12. anonymous

Is this correct? cos(y) (1) + x (-sin(y) y' = 0 cos(y) - x sin(y) y' = 0 cos(y) = x sin(y) y' $y' = \frac{ \cos(y) }{ x \sin (y) }$ $y' = \frac{ \cos(y) }{ x \sin (y) } = \frac{ \cos \pi }{ (-1) \sin \pi } = \frac{ -1 }{ (-1)(0) }$ Hence it is undefined

13. dessyj1

The math looks correct, but I do not think it would be undefined, rather it would not exist.

14. anonymous

@dessyj1 Thank you

15. dessyj1

np