Question

x^2 cos^2 y - sin y = 0 find the derivative, I'm seriously having trouble with this problem, any help would be appreciated.

Answer 1

\[2x\cos^2( y)+x^22\cos(y)(-\sin(y))y' - \cos(y)y' = 0\]

solve for y'

Answer 2

how did you get there?
this is where I am at
\[x ^{2}*-2cosy siny+2xcos ^{2}y -\cos y=0\]

Answer 3

I assume you are doing implicit differentiation ... therefore every time you differentiate y with respect to x you get y'

or \[\frac{dy}{dx}\]

Answer 4

I am, just having trouble understand how you got from A to B. Your syntax is just confusing me.

Answer 5

understanding*

Answer 6

you have the same thing as me except you are missing the y'

Answer 7

okay, well this is basically my question then, how do i differentiate cos^2 y and sin y

Answer 8

treat every y as y(x) (it is a function of x) so when you take the derivative you need to use the chain rule

Answer 9

\[\frac{d}{dx}\sin(y)=\cos(y)\frac{dy}{dx}\]

Answer 10

so
\[d/dx(\cos^2 y) \neq -2\cos y \sin y\]

Answer 11

\[\frac{d}{dx}(\cos^2(x))=-2\cos(x)\sin(x)\]

but

\[\frac{d}{dx}(\cos^2(y))=-2\cos(y)\sin(y)\frac{dy}{dx}\]

Answer 12

\[y'=\frac{2 x \cos (y)}{2 x^2 \sin (y)+1} \]

Answer 13

could you just show me the steps of getting from cos^2 y to - 2 cos y sin y dy/dx

Answer 14

bring down the 2 ...keep the inside the same (the cosine) ...drop the power down by 1...multiply by the derivative of the inside ...derivative of cosine is -sine...keep the inside of that the same (which is y)...then again by the chain rule multiply by the derivative of the inside...the derivative of y is dy/dx

Answer 15

bring down the 2?

Answer 16

\[\frac{d}{dx}(f(x))^2=2(f(x))^1f'(x)\]

bring the two from the exponent ... out to the front.

Answer 17

The attachment show how I solved it.

Answer 18

...?

Answer 19

you could also let \[F(x,y)=x^2 \cos^2 y - \sin y \]

and compute

\[-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\]

;)

Answer 20

okay I guess my misunderstanding is how to apply the chain rule to a function with a y with in it. I have been working on the same problem for the past hour, this is in no way efficient learning.

(Cos y)^2

lets say that y = u^2
u = cos m
and
m = y

lets also say that

y prime = 2 u
u prime = -sin m
and
m prime = dy/dx

the chain rule states that

the derivative of f(g(x)) = fprime( g(x)) * gprime of (x)

so 2 (cos m) * - sin (m)
thats just the chain rule
now how do I apply the last m ?

Your explanations are a little vague for me to understand, I learn by example, if you were to just break it down to a walkthrough that would be fantastic, btw thanks for your time.

Answer 21

\[\cos^2(y)=\cos^2(y(x))\]

\[\frac{d}{dx}\cos^2(y(x))=2(\cos(y(x))^1(-\sin(y(x))y'(x))\]

\[=-2\cos(y)\sin(y)y'\]

Answer 22

why is there a superscript 1?

Answer 23

I don't understand why this is confusing me so much. :(

Answer 24

well, I give up.... I suppose I'll give you a medal for your effort.... :(