Question

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.

Answer 1

So start with that product on the left side.

Answer 2

huh?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

im confused!

Answer 8

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

Answer 9

I love how errbody helps him and no one else.

Answer 10

oh.

Answer 11

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

Answer 12

my bad.

Answer 13

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

Answer 14

is that correct? Please help?

Answer 15

How did you differentiate (cosy)^2?

Answer 16

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

Answer 17

...

Answer 18

\[x^2 Cos[y]^2 - Sin[y] == 0 \]
\[Cos[y]^2 = 1 - Sin[y]^2 \]
\[x^2 (1-\text{Sin}[y]{}^{\wedge}2)-\text{Sin}[y]==0 \]
\[x^2-\text{Sin}[y]-x^2 \text{Sin}[y]^2=0 \]
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.

Answer 19

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