Question about derivatives

http://www.teacherschoice.com.au/images/derivative_definition_formula.gif

What I don't understand is why do we consider x as a constant and h as a variable? How can we be sure that it's true for all x? Is it better to think of it as the derivative of f(t) at t=x?

Reply 1

a fresh student

gangsta316

http://www.teacherschoice.com.au/images/derivative_definition_formula.gif

What I don't understand is why do we consider x as a constant and h as a variable? How can we be sure that it's true for all x? Is it better to think of it as the derivative of f(t) at t=x?

It is just the opposite - x is the variable, and h is the constant.
h is a very small number tending to 0.
This is the definition of derivative, which can be undestood as a gradient over an interval h tendng to 0.
delta y = f(x+h)-f(x);
delta x = h

Reply 2

AsakuraMinamiFan

gangsta316

http://www.teacherschoice.com.au/images/derivative_definition_formula.gif

What I don't understand is why do we consider x as a constant and h as a variable? How can we be sure that it's true for all x? Is it better to think of it as the derivative of f(t) at t=x?

It depends on context. By definition, differentiability is only one one point. We say it holds on a set (such as a < x < b which we may write as (a,b)) if it is differentiable at every x such that x is in this set (so for example, x is in (a,b)).

The difference between differentiability at a point and differentiability on a set is that the first one is only one value of the limit, the second one is a set of values of the limit. So we can define a function from the set of differentiable points to the set of values of the limit.

Reply 3

birdsong1

We find the derivative at a fixed x, and we do this by taking the limit as h tends to 0.

Reply 4

john !!

14

x is not fixed! neither is h.

h is what you call a dummy variable, it appears on the right hand side but not the left, so neither side must depend on it necessarily, (sort of like the summation variable), but it does vary. The derivative needs a neighbourhood around the point to be defined, and h characterises this neighbourhood.

Reply 5

gangsta316

OP

13

The prof. gave two other similar definitions of the derivative before giving us this familiar one.

The derivative of f(x) at x = a:

lim x->a

\displaystyle\frac{f(x)-f(a)}{x-a}

This makes sense because it's "change in y/change in x".

f(x) is differentiable at x=a if

lim h-> 0

\displaystyle\frac{f(a+h)-f(a)}{h}

exists.

This definition
http://www.teacherschoice.com.au/images/derivative_definition_formula.gif

seems to be the second definition but at x = x. So isn't it treating x, f(x) as a fixed point like a and f(a)? Is that allowed?

Reply 6

birdsong1

john !!

x is not fixed! neither is h.

That's a matter of philosophy. In this case, it's easier for the OP to think of us computing an expression for arbitrary fixed x - but of course, because x was arbitrary, we've managed to compute it for all x.

Reply 7

birdsong1

gangsta316

The prof. gave two other similar definitions of the derivative before giving us this familiar one.

The derivative of f(x) at x = a:

lim x->a

\displaystyle\frac{f(x)-f(a)}{x-a}

This makes sense because it's "change in y/change in x".

They just named the variables differently. Old x is a here, and old x + h is x here.

gangsta316

f(x) is differentiable at x=a if

lim h-> 0

\displaystyle\frac{f(a+h)-f(a)}{h}

exists.

This definition
http://www.teacherschoice.com.au/images/derivative_definition_formula.gif

seems to be the second definition but at x = x. So isn't it treating x, f(x) as a fixed point like a and f(a)? Is that allowed?

Similarly, variables were just renamed. Old x is a here, old h is still new h.

Question about derivatives

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