Implicit Differentiation

  • Thread starter mlowery
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  • #1
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Find [tex]\frac{dy}{dx}[/tex] given [tex]cos(y^2) = x^4[/tex]
Is this correct:

1. [tex]cos(y^2) = x^4[/tex]

2. [tex]-sin(y^2) \times 2y \frac{dy}{dx} = 4x^3[/tex]

3. [tex]\frac{dy}{dx} = \frac{4x^3}{-2sin(y^2)}[/tex]
 
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Answers and Replies

  • #2
Doc Al
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Oops... Looks like you left out a "y" in the last step. Write it like this:
[tex]\frac{dy}{dx} = -\frac{2x^3}{y \sin(y^2)}[/tex]
 
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  • #3
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Yes, that is what I originally did. The thing is, this is a multiple choice question. Of the choices, the only answers close to this are: (Please don't think I am using this forum for answers. I just believe none of the choices are correct).

A) [tex]\frac{4x^3}{-sin(y^2)}[/tex]

B) [tex]\frac{4x^3}{-2ysin(y^2)}[/tex]

Here are the other choices:

C) [tex]\frac{\sqrt{xy}-y}{2xy}[/tex]

D) [tex]\frac{x^4}{-sin(y^2)}[/tex]

E) [tex]\frac{4x^3}{cos(2y)}[/tex]
 
  • #4
mezarashi
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For the second step, I'd prefer having it written as:

[tex]-sin(y^2) \times 2y dy = 4x^3dx[/tex]

This will probably be more consistent once you encounter more complicated problems or do multivariable calculus.

Answer B is correct. You forgot to move the y over in your last step, step 3.
 
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  • #5
Doc Al
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mezarashi said:
Answer B is correct. You forgot to move the y over in your last step, step 3.
Right! (I just realized that you left out that y in your last step!)
 
  • #6
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Hehe, how'd that "y" sneak by me? :grumpy:
Thanks for the help.

Yeah mezarashi, I am not quite familiar with the notation you used (I just started using the dy/dx notation last week).

Thanks,
Mitch
 

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