Equation of a Tangent to a curve

Question

$ f' (x) dx = dy $

Would I be correct in assuming that this is the equation of any tangent line to the curve f (x) ?

Answer 1

No, to get the equation of a tanget line, you need to evaluate the derivative of a curve (at a certain point x0). Therefore, the slope of the tangent line is: $$f'(x_o)=m$$ you get the equation of the tangent line by plugging the point: $$P(x_0, f(x_0))$$ to: $$y-y_0=m(x-x_0)$$ and therefore you get: $$y = f'(x_0)m-f'(x_0)x_0+f(x_0)$$

Answer 2

"No, to get the equation of a tanget line, you need to equal the derivative of a curve (at a certain point x0) to zero. Therefore, the slope of the tangent line is: f′(xo)=0."

Where did you get that idea? The slope of the tangent linear at $(x_0, f(x_0))$ is $f'(x_0)$. That is not necessarily 0 except where the tangent line is horizontal (which will happen at max or min points- perhaps you are confusing this with the problem of finding max and min.)

Answer 3

I disagree with the "no"'s. You can think of $dx$ and $dy$ as new variables related by $dy = f'(x) dx.$ If you place the origin of the $dx-dy$ plane on the point $(x_0,f(x_0))$, then $dy=f'(x_0) dx$ is indeed the equation of that line (in $dx-dy$ coordinates.) It's a good way for beginners to think of differentials, because it leads naturally to linear approximations.