Understanding Derivatives: Finding the Derivative of y=2x^2-5

In summary, Dan was working on a problem but was having trouble figuring out what he was doing wrong. He was eventually able to figure it out and thanked the people who helped him.
  • #1
chrisdapos
23
0
Hello, today my teacher presented us with the concept of derivatives. I was okay in class, but when I got home, I was completely lost. Using the definition (f(x+h) - f(x))/h, we have to fine the derivative of y=2x^2 - 5. I know the answer is 4x from the back of the book, but I can't understand how to get it. Am I factoring something wrong? I get to a point of simplification that everything just cancels out. Your help is greatly appriciated!
 
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  • #2
The definition you want to use is this:

[tex]\lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}[/tex]

Now, set [itex]f(x)=2x^2-5[/itex] and insert this in the expression, simplify and take the limit.
 
  • #3
Quick check: When you expand the numerator, f(x+h)-f(x), you should get an expression that cancels the h in the denominator.

-Dan
 
  • #4
my problem is when i simplify, i get 0 over 0...i don't know why...i can't get the h to go away in the denominator
 
  • #5
chrisdapos said:
my problem is when i simplify, i get 0 over 0...i don't know why...i can't get the h to go away in the denominator

[tex] f(x+h)=2(x+h)^2-5=2x^2+4xh+2h^2-5 [/tex]
and
[tex] f(x)=2x^2-5 [/tex]
So when you subtract the two you get...

-Dan
 
Last edited:
  • #6
Then you must be doing something wrong. Remember this: If [itex]f(x)=2x^2-5[/itex], what is then [itex]f(x+h)[/itex]? Check your calculations again.
 
  • #7
Thank you so much guys! Wow i feel dumb...turns out i was just plugging the functions in wrong. Thank you so much!
 
  • #8
chrisdapos said:
Thank you so much guys! Wow i feel dumb...turns out i was just plugging the functions in wrong. Thank you so much!

Hey, as I always tell my students: it's only a dumb question when you DON'T ask it! :biggrin:

-Dan
 

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