Asked by lil
what is the derivative of sinx- cosy = 0 in its simplest form? I got to the point y'=-consx/siny, but I wasn't too sure. This could probably simplified further. help!
You're supposed to treat y as a function of x and differentiate it implicitly. We have
d/dx{sinx- cosy)=d/dx{sinx) - d/dx{cosy)=
cos(x)-(-sin(y)*y')=0
You should be able to solve for y'.
I did, and I ended up with y'=-cos(x)/sin(y)
How do I simplify this further?
That's as far as you can go unless you know something specific about y. Without knowing something about y(x) there's nothing else to do here.
You're supposed to treat y as a function of x and differentiate it implicitly. We have
d/dx{sinx- cosy)=d/dx{sinx) - d/dx{cosy)=
cos(x)-(-sin(y)*y')=0
You should be able to solve for y'.
I did, and I ended up with y'=-cos(x)/sin(y)
How do I simplify this further?
That's as far as you can go unless you know something specific about y. Without knowing something about y(x) there's nothing else to do here.
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