## 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.

1. anonymous

2. anonymous

huh?

3. anonymous

Use the product rule to deal with x^2(cosy)^2

4. anonymous

dont i need to use chain rule before I can apply the product rule?

5. anonymous

Chain rule comes up in the process of applying the product rule when you differentiate the cosine term.

robtobey, take a look at that pigpen problem and let me know if I made an error

7. anonymous

im confused!

8. 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

9. anonymous

I love how errbody helps him and no one else.

10. anonymous

oh.

11. anonymous

so I got: 2y (dy/dx) (cosy) (-siny) (dy/dx) by the chain rule. is that correct?

12. anonymous

13. anonymous

$2x \cos^{2}y+x^{2} 2y (dy/dx) (cosy) (-\sin) (dy/dx)$

14. anonymous

15. anonymous

How did you differentiate (cosy)^2?

16. anonymous

2y (dy/dx) (cosy) (-siny) (dy/dx)

17. anonymous

...

18. 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.

19. 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