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
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

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
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.