1. ## basic question about designation

I am learning calculus at home during my summer break.In my current lesson, here's the problem:Let 2x squared y + y squared = cos x.Find dy/dx.The solution is dy/dx = -sinx -4xy/2x squared + 2yThe intermediate steps:2x^2 dy/dx + y(4x) + 2y dy/dx = -sin xdy/dx(2x^2 + 2y)= -sin x - 4xyWhy is there dy/dx only for 2x squared and 2y and not -sin x and y(4x)?Any light you could shed on this would be greatly appreciated.Carrie

2. Hello! Please visit this thread to learn how to write posts using LaTeX: http://www.mathhelpforum.com/math-he...-tutorial.html

As your post is at the moment, it's very hard to figure out what the question is.

3. Yes, LaTeX is very fun to use. I think you'll like it.

The reason we don't get a factor of $\displaystyle \frac{dy}{dx}$ is that the Product Rule states

$\displaystyle \frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x).$

Here, $\displaystyle f(x)=2x^2$ and $\displaystyle g(x)=y$. Thus, we have

$\displaystyle \frac{d}{dx}(2x^2\cdot y)=4x\cdot y + 2x^2\cdot\frac{dy}{dx}.$

4. ## Thank you

Yes, I see. Thanks for your time, and I"ll check the formatting rules for future use. Please forgive me.
Carrie