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.

`d/dx(2x-5x^3y^2+4y=0)`

`2-15x^2y^2-10x^3yy'+4y'=0`

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

`-10x^3yy'+4y'=15x^2y^2-2`

`y'(4-10x^3y)=15x^2y^2-2`

`y'=(15x^2y^2-2)/(4-10x^3y)`

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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