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

[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?

 

[HallsofIvy]

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

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

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?

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.

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

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.