Notation Question

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
This is about calculus notation, not taking a derivative.

1#

\(\displaystyle f'(x) = \dfrac{d}{dx}[x^{2}] - \dfrac{d}{dx}[x]\) - I know this is right.

OR


# 2

\(\displaystyle f'(x) = \dfrac{dy}{dx}[x^{2}] - \dfrac{dy}{dx}[x]\) :confused:

OR


# 3

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{dx}[x^{2}] - \dfrac{dy}{dx}[x]\) :confused:

OR

# 4

\(\displaystyle \dfrac{dy}{dx} = \dfrac{d}{dx}[x^{2}] - \dfrac{d}{dx}[x]\) :confused:
 
Last edited:
This is about calculus notation, not taking a derivative.

1#

\(\displaystyle f'(x) = \dfrac{d}{dx}[x^{2}] - \dfrac{d}{dx}[x]\) - I know this is right.

OR


# 2

\(\displaystyle f'(x) = \dfrac{dy}{dx}[x^{2}] - \dfrac{dy}{dx}[x]\) :confused:

OR


# 3

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{dx}[x^{2}] - \dfrac{dy}{dx}[x]\) :confused:

OR

# 4

\(\displaystyle \dfrac{dy}{dx} = \dfrac{d}{dx}[x^{2}] - \dfrac{d}{dx}[x]\) :confused:
If #1 is preceded by the function definition, \(\displaystyle f(x) = x^2 - x\), it is ok.

If #4 is preceded by the function definition, \(\displaystyle y = x^2 - x\), it is ok.

The other two show (x^2 - x) being multiplied by a function (dy/dx). Not at all related to the operator (d/dx) operating on (x^2 - x).

\(\displaystyle \dfrac{\mathrm d}{\mathrm d x}\) is an operator

\(\displaystyle \dfrac{\mathrm d y}{\mathrm d x}\) is the function resulting when \(\displaystyle \dfrac{\mathrm d}{\mathrm d x}\) has operated on \(\displaystyle y\).
 
If #1 is preceded by the function definition, \(\displaystyle f(x) = x^2 - x\), it is ok.

If #4 is preceded by the function definition, \(\displaystyle y = x^2 - x\), it is ok.

The other two show (x^2 - x) being multiplied by a function (dy/dx). Not at all related to the operator (d/dx) operating on (x^2 - x).

\(\displaystyle \dfrac{\mathrm d}{\mathrm d x}\) is an operator

\(\displaystyle \dfrac{\mathrm d y}{\mathrm d x}\) is the function resulting when \(\displaystyle \dfrac{\mathrm d}{\mathrm d x}\) has operated on \(\displaystyle y\).

ok, thanks.
 
Top