Question

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

Answer 1

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)

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

product rule, kid

Answer 6

@nincompoop I know the product rule

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Remember that the derivative of a constant is 0

Answer 12

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

Answer 13

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

Answer 14

@dessyj1 Thank you

Answer 15

np