Evaluate the derivitive dy/dx at the (0,0) for the equation


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You simply differentiate whole expression, but when you differentiate `y` you need to remember that `y` is a function of `x`  and thus `(f(y))'=f'(y)y'` in other words when you have some function of `y` you differentiate it as if it's a composition i.e. you use chain rule.

E.g. `(y^2)'=2yy'`

Let's now differentiate our equation.



Now leave everything with `y'` on left side and everything else on the right side.




Hence `dy/dx=(15x^2y^2-2)/(4-10x^3y)`

Now to calculate the value of the derivative at (0,0) we simply put zeros instead of x and y.

`dy/dx(0,0)=(15cdot0cdot0-2)/(4-10cdot0cdot0)=-2/4=-1/2` <-- Your solution

In the attached image blue is the curve defined by your equation and red is the tangent line at point (0,0).

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