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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.
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So start with that product on the left side.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the product rule to deal with x^2(cosy)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont i need to use chain rule before I can apply the product rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Chain rule comes up in the process of applying the product rule when you differentiate the cosine term.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0robtobey, take a look at that pigpen problem and let me know if I made an error

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Derivative 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
 5 years ago
Best ResponseYou've already chosen the best response.0I love how errbody helps him and no one else.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I got: 2y (dy/dx) (cosy) (siny) (dy/dx) by the chain rule. is that correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2x \cos^{2}y+x^{2} 2y (dy/dx) (cosy) (\sin) (dy/dx)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that correct? Please help?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you differentiate (cosy)^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02y (dy/dx) (cosy) (siny) (dy/dx)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2 Cos[y]^2  Sin[y] == 0 \] \[Cos[y]^2 = 1  Sin[y]^2 \] \[x^2 (1\text{Sin}[y]{}^{\wedge}2)\text{Sin}[y]==0 \] \[x^2\text{Sin}[y]x^2 \text{Sin}[y]^2=0 \] Let u equal Sin[y] \[u+x^2u^2 x^2=0 \] 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
 5 years ago
Best ResponseYou've already chosen the best response.0Think 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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