- #1
seand
- 7
- 0
Hi there, I'm a new user to the forums (and Calculus) and I 'm hoping you can give me your opinion on my chain rule form below. When learning the chain rule, I was taught two forms. This form:
[tex]\frac{d}{dx}f(g(x))=f'(g(x))g'(x)[/tex]
As well as the Leibniz form
[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/tex] where [tex]y=f(u)[/tex] and [tex]u=g(x)[/tex]
I prefer the Leibniz notation, except that it requires you to understand that [tex]y=f(u)[/tex] and [tex]u=g(x)[/tex], to really understand the d/dx expression.
So my question is if there is a way to make Liebniz more explicit? ie. Does the following make sense? Is it correct?
[tex]\frac{d}{dx}f(g(x)) = \frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}[/tex]
[tex]\frac{d}{dx}f(g(x))=f'(g(x))g'(x)[/tex]
As well as the Leibniz form
[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/tex] where [tex]y=f(u)[/tex] and [tex]u=g(x)[/tex]
I prefer the Leibniz notation, except that it requires you to understand that [tex]y=f(u)[/tex] and [tex]u=g(x)[/tex], to really understand the d/dx expression.
So my question is if there is a way to make Liebniz more explicit? ie. Does the following make sense? Is it correct?
[tex]\frac{d}{dx}f(g(x)) = \frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}[/tex]