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anonymous

  • 5 years ago

How do I solve \[x^{2} \cos ^{2} y -siny = 0\] ? Can somebody guide me through the steps. I know its implicit differentiation. I just need a bit of guidance.

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  1. anonymous
    • 5 years ago
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    So start with that product on the left side.

  2. anonymous
    • 5 years ago
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    huh?

  3. anonymous
    • 5 years ago
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    Use the product rule to deal with x^2(cosy)^2

  4. anonymous
    • 5 years ago
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    dont i need to use chain rule before I can apply the product rule?

  5. anonymous
    • 5 years ago
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    Chain rule comes up in the process of applying the product rule when you differentiate the cosine term.

  6. radar
    • 5 years ago
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    robtobey, take a look at that pigpen problem and let me know if I made an error

  7. anonymous
    • 5 years ago
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    im confused!

  8. anonymous
    • 5 years ago
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    Derivative of the first x^2 times the second (Cosy)^2 + x^2 times the derivative of (cosy)^2.... you would use the chain rule to find the derivative of (Cosy)^2

  9. anonymous
    • 5 years ago
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    I love how errbody helps him and no one else.

  10. anonymous
    • 5 years ago
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    oh.

  11. anonymous
    • 5 years ago
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    so I got: 2y (dy/dx) (cosy) (-siny) (dy/dx) by the chain rule. is that correct?

  12. anonymous
    • 5 years ago
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    my bad.

  13. anonymous
    • 5 years ago
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    \[2x \cos^{2}y+x^{2} 2y (dy/dx) (cosy) (-\sin) (dy/dx)\]

  14. anonymous
    • 5 years ago
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    is that correct? Please help?

  15. anonymous
    • 5 years ago
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    How did you differentiate (cosy)^2?

  16. anonymous
    • 5 years ago
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    2y (dy/dx) (cosy) (-siny) (dy/dx)

  17. anonymous
    • 5 years ago
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    ...

  18. anonymous
    • 5 years ago
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    \[x^2 Cos[y]^2 - Sin[y] == 0 \] \[Cos[y]^2 = 1 - Sin[y]^2 \] \[x^2 (1-\text{Sin}[y]{}^{\wedge}2)-\text{Sin}[y]==0 \] \[x^2-\text{Sin}[y]-x^2 \text{Sin}[y]^2=0 \] Let u equal Sin[y] \[-u+x^2-u^2 x^2=0 \] Solve the above for u \[\left\{u\to \frac{-1-\sqrt{1+4 x^4}}{2 x^2}\right\},\left\{u\to \frac{-1+\sqrt{1+4 x^4}}{2 x^2}\right\} \] Replace u with Sin[y] \[\left\{\text{Sin}[y]\to \frac{-1-\sqrt{1+4 x^4}}{2 x^2}\right\},\left\{\text{Sin}[y]\to \frac{-1+\sqrt{1+4 x^4}}{2 x^2}\right\} \] \[y==\text{ArcSin}\left[\frac{-1-\sqrt{1+4 x^4}}{2 x^2}\right],y==\text{ArcSin}\left[\frac{-1+\sqrt{1+4 x^4}}{2 x^2}\right] \] I hope there no mistakes.

  19. anonymous
    • 5 years ago
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    Think of it as (cos(f(x))^2 where y=f(x). That first y dy/dx shouldnt come out So you get 2(cos(f(x))(-sin(f(x))f'(x). I believe

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