# Simple question -definition of derivative

• Mar 7th 2010, 01:26 PM
pinkperil
Simple question -definition of derivative
We've all seen the definition, but for some reason I still don't understand the logic behind it. This is the part of the definition that I have a problem with:

f (x+h) - f (x) / (x+h) - x

I understand that (x+h) - x = h.

If this is the case, will this be true f (x+h) - f (x) = h ...?

Thus, h / h ?
• Mar 7th 2010, 01:49 PM
pickslides
Hi there,

This part is given

$\displaystyle \frac{f (x+h) - f (x)}{(x+h) - x}$

then as you have shown there is some cancellation in the denominator

$\displaystyle \frac{f (x+h) - f (x)}{h}$

This is where it stops. No more work needs to be done.

If you want to find a derivative from here find the limit of your function as $\displaystyle h \to 0$ using the deifnition above.
• Mar 7th 2010, 02:10 PM
Plato
Quote:

Originally Posted by pinkperil
We've all seen the definition, but for some reason I still don't understand the logic behind it.

Do you understand how to determinte the slope of the chord between the two points $\displaystyle \left(x,f(x)\right)~\&~\left(x+h,f(x+h)\right)?$
That is the logic of it all.
• Mar 7th 2010, 02:20 PM
HallsofIvy
Quote:

Originally Posted by pinkperil
We've all seen the definition, but for some reason I still don't understand the logic behind it. This is the part of the definition that I have a problem with:

f (x+h) - f (x) / (x+h) - x

I understand that (x+h) - x = h.

If this is the case, will this be true f (x+h) - f (x) = h ...?

NO, f(x+ h) is NOT equal to f(x)+ h! Exactly what it is depends on the function f.

Quote:

Thus, h / h ?
Well, that would make calculus pretty trivial wouldn't it?

If f(x) is a linear function, f(x)= ax+ b, then f(x+h)= a(x+h)+ b= ax+ah+ b.
f(x+h)= ax+ah+ b- ax+b= ah. (f(x+h)- f(x))/h= ah/h= a, the slope of the line.

If $\displaystyle f(x)= ax^2+ bx+ c$, then $\displaystyle f(x+h)= a(x+h)^2+ b(x+h+ c$$\displaystyle = ax^2+ 2axh+ h^2+ bx+bh+ c$ so $\displaystyle f(x+h)- f(x)= ax^2+ 2axh+ h^2+ bx+ bh+ c- ax^2- bx- c= 2axh+ h^2+ bh$ and the "difference quotient" $\displaystyle \frac{f(x+h)- f(x)}{h}= \frac{2axh+ h^2+ bh}{h}= 2ax+ h+ b$

It gets really complicted for things like sin(x), cos(x), or log(x)!
• Mar 7th 2010, 03:53 PM
pinkperil
Quote:

Originally Posted by HallsofIvy
NO, f(x+ h) is NOT equal to f(x)+ h! Exactly what it is depends on the function f.

I see where my problem is. My problem is something even more fundamental: misunderstanding of functions. So if the function isn't what I assumed, then I must assume this:

y = f(x) = 9

f(x+h) - f(x) / h

9 - 9 / h

Although, it doesn't make much sense to me how f(x + h) = f(x) when this isn't true in many other cases like this:

f(x) = 2x
f(x + h) = 2x + 2h

Is it because the first situation has only a constant and the other has a variable? The reason why I thought f(x+h) - f(x) = h is because: (constant + h) - constant = h.
• Mar 8th 2010, 04:02 AM
HallsofIvy
Quote:

Originally Posted by pinkperil
I see where my problem is. My problem is something even more fundamental: misunderstanding of functions. So if the function isn't what I assumed, then I must assume this:

y = f(x) = 9

f(x+h) - f(x) / h

9 - 9 / h

Although, it doesn't make much sense to me how f(x + h) = f(x) when this isn't true in many other cases like this:

You quoted part of my post so you must have read it. How could you get "f(x+h)= f(x)" from it? f(x+h)= f(x) only for a constant function such as "f(x)= 9" where f(x)= f(y) no matter what x and y are.

And why would you say "if the function isn't what I assumed". You did not mention any particular function in your original post.

Quote:

f(x) = 2x
f(x + h) = 2x + 2h

Is it because the first situation has only a constant and the other has a variable? The reason why I thought f(x+h) - f(x) = h is because: (constant + h) - constant = h.
If f is a constant, f(x)= c, then f(x+h)= c also so f(x+h)- f(x)= 0 and then (f(x+h)- f(x))/h= 0. Of course, the graph of a constant function is a horizontal straight line and has slope 0.

The function f(x)= x+ c has f(x+h)= x+ h+ c so f(x+ c)- f(x)= x+h+c- x- c= h and then (f(x+h)- f(x))/h= 1. The line y= x+ c has slope 1.

IF you were refering to specific function you should have said so in your first post. It appeared that your "f(x)" could be any function.
• Mar 8th 2010, 12:41 PM
pinkperil
Quote:

...only for a constant function such as "f(x)= 9" where f(x)= f(y) no matter what x and y are.
This makes sense to me now. I couldn't wrap my head around (x + h) being something whole and my mind kept wanting to break it apart. Thanks a lot for your help. I appreciate it.