anonymous
  • anonymous
Find dy/dx of x cos(y) = 1 at the point (-1,pi) using implicit differentiation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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)
dessyj1
  • dessyj1
It seems like we will have to use the product rule. Are you familiar with the product rule?
nincompoop
  • nincompoop
do you want to just get the general derivative implicitly first then add the values later

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anonymous
  • anonymous
Yes. But do I keep the x as it is or do I rplacee it with -1?
nincompoop
  • nincompoop
product rule, kid
anonymous
  • anonymous
@nincompoop I know the product rule
dessyj1
  • dessyj1
Do you need help with implicitly deriving the left side of your equation?
anonymous
  • 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
dessyj1
  • dessyj1
It would not be cos(y)(0) because the derivative of x is 1
dessyj1
  • dessyj1
Also, of you were following the f'g +fg' you will realize that you forgot to multiply y'sin(y) by an x.
dessyj1
  • dessyj1
Remember that the derivative of a constant is 0
anonymous
  • 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
dessyj1
  • dessyj1
The math looks correct, but I do not think it would be undefined, rather it would not exist.
anonymous
  • anonymous
@dessyj1 Thank you
dessyj1
  • dessyj1
np

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