What is done in the second line of the product rule proof?

[Muthumanimaran]

What has done here in the second line of the proof for product rule?, from Mathematical methods for physicists from Riley, Hobson
they defined f(x)=u(x)v(x) and these steps are given,
I have no idea how to proceed further please help me.

 

[Simon Bridge]

Do you know why they bothered to work out $f(x+\Delta x) - f(x)$ at all?
I mean: what's the point?

How would you normally go about proving the product rule - if you didn't have the example from Riley and Hobson?
i.e. do you know the definition of the derivative?

 

[Muthumanimaran]

Do you know why they bothered to work out f(x+Δx)−f(x)f(x+\Delta x) - f(x) at all?
I mean: what's the point?

I don't know

How would you normally go about proving the product rule - if you didn't have the example from Riley and Hobson?
i.e. do you know the definition of the derivative?

yes, derivative is the rate of one function to another function, it actually says how fast one function changes with respect to other, am I right?

 

[Simon Bridge]

yes, derivative is the rate of one function to another function, it actually says how fast one function changes with respect to other, am I right?

Not exactly ... that was the description of what the derivative is, not the definition. The definition is: $$f^\prime(x) = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$$ ... this gives the derivative of f(x) with respect to x.

To prove the product rule, first set $f(x)=v(x)u(x)$ then apply the definition to f.

 

[Muthumanimaran]

Not exactly ... that was the description of what the derivative is, not the definition. The definition is: $$f^\prime(x) = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$$ ... this gives the derivative of f(x) with respect to x.

To prove the product rule, first set $f(x)=v(x)u(x)$ then apply the definition to f.

got it. Thank you