anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
So start with that product on the left side.
anonymous
  • anonymous
huh?
anonymous
  • anonymous
Use the product rule to deal with x^2(cosy)^2

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anonymous
  • anonymous
dont i need to use chain rule before I can apply the product rule?
anonymous
  • anonymous
Chain rule comes up in the process of applying the product rule when you differentiate the cosine term.
radar
  • radar
robtobey, take a look at that pigpen problem and let me know if I made an error
anonymous
  • anonymous
im confused!
anonymous
  • anonymous
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
anonymous
  • anonymous
I love how errbody helps him and no one else.
anonymous
  • anonymous
oh.
anonymous
  • anonymous
so I got: 2y (dy/dx) (cosy) (-siny) (dy/dx) by the chain rule. is that correct?
anonymous
  • anonymous
my bad.
anonymous
  • anonymous
\[2x \cos^{2}y+x^{2} 2y (dy/dx) (cosy) (-\sin) (dy/dx)\]
anonymous
  • anonymous
is that correct? Please help?
anonymous
  • anonymous
How did you differentiate (cosy)^2?
anonymous
  • anonymous
2y (dy/dx) (cosy) (-siny) (dy/dx)
anonymous
  • anonymous
...
anonymous
  • anonymous
\[x^2 Cos[y]^2 - Sin[y] == 0 \] \[Cos[y]^2 = 1 - Sin[y]^2 \] \[x^2 (1-\text{Sin}[y]{}^{\wedge}2)-\text{Sin}[y]==0 \] \[x^2-\text{Sin}[y]-x^2 \text{Sin}[y]^2=0 \] 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.
anonymous
  • anonymous
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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