# Using differentials in the equation of tangent line to the curve x^2 at point 3

1. Jan 14, 2010

### Juwane

To understand differentials better, I'm trying to use differentials dy and dx in the equation of the tangent line to the curve x^2 at point 3.

Here is the equation of the tangent line to the curve x^2 at point 3:

$$y=f'(3)(x-3)+f(3)=2(3)(x-3)+9=6(x-3)+9$$

But since we are dealing with the tangent, why can't we write the above equation as:

$$y=\frac{dy}{dx}(x-3)+dy$$

Since $$\frac{dy}{dx}=6$$ from which $$dy=6dx$$, we can rewrite the above as:

$$6(x-3)+6dx=6(x-3)+6(3)=6(x-3)+18$$

Why this equation and the first equation aren't the same?

How else can the equation of tangent line be written in terms of dys and dxs?

Last edited: Jan 14, 2010
2. Jan 15, 2010

### HallsofIvy

Staff Emeritus
The dy/dx in place of f' is fine but dy is NOT 9!

Again, it is because the added "9" is NOT dy and so not "6dx". You are not distinguishing between "dy/dx" and "$\Delta y/\Delta x$"
$$\frac{dy}{dx}= \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$$
You can't just ignore the limit.

Well, if y= (dy/dx)(x- a)+ y(a) then dx(y- y(a))= dy(x-a). Does that help? It is at least more "symmetric" than the usual form.