Calculating the derivative of x^2 using the definition of the derivitave

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Category:

Education

Tags:

• derivative
• x^2
• calculus
• limit
• as
• approaches
• zero
• calc

[MrFalcon675]

I believe that you are supposed to cancel out the h's in the numerator and denominator before you make the approximation that h~0, otherwise 2hx would go to zero.

[barbarossaaaa]

@MrFalcon675

remember that h is a value, that is very close to, but never actually zero, so that for the purposes of calculating the instantaneous rate of change we can say that

2(0.0000000000001)(x) ≈ 2(0)(x) so h is never actually zero and wouldn't cause 2hx to result in zero due to multiplying by zero

[mikemustmurder]

Um.. Dude the h's are supposed to cancel eachother out. If you just said they were equal to zero it would be 0/0

[barbarossaaaa]

@mikemustmurder

true but then youd still be left with h^2/h = h  ... i wanted to emphasize that the interval of f(x+h)-f(x) is small enough so that it could be considered an instantaneous rate of change

[ApollosInsight]

Barbarosa, what are you going to school for to use this math?

[barbarossaaaa]

@ApollosInsight

chemical engineering, but the practical applications of this process are boundless.

[tomtomvader]

the last step should actually be two steps. first you factor out an h on the top... h(2x + h) and cancel it out with the h on the bottom. then your left with 2x + h, in which you can now substitute zero for the h which gives you 2x. the reason you couldnt make the substitution when the h was on the bottom is because it would make the denominator zero, which is a no no (this coming from a physics major and an mra) But the way you talk about math shows real promise. keep it up barb

[fomiz]

You need to consider teaching! 

[Mathman1ac]

@didles123

Incorrect. The change you are talking about above is algebra. In other words, if you didn't perform a differentiation, you are not using l'Hospital's rule.

[didles123]

@Mathman1ac l'Hospital's rule changes 2hx + h^2 / h into 2x + 2h which makes it obvious that it is 2x as the limit goes to zero. It's similar to how dividing both sides by h gives 2x + h. l'Hospital's rule helps you solve limits, MrFalcon was talking about limits.

[Mathman1ac]

@didles123

Not really. l'Hospital's rule is ONLY applicable in determining if a continuous function has a derivitive at an arbitrary point on it's domain. It doesn't really help you FIND the derivitive because you still have to apply standard rules of differentiation just to use it. Heck, it doesn't even help you find the domain of the derivitve. It's almost like a tool you use to make sure you don't waste your time with an insolluble problem.

[thinfoxgmail]

@barbarossaaaa i think he meant that you can simplify it a bit further to say:

(2hx + h^2)/h => h(2x + h)/h => 2x + h =>2x as h->0

[sillygoose544]

I'm taking Calculus next semester (being a former liberal arts major, I never thought I'd be saying this.) I just started studying limits now, and I cant wait! (something else I never thought I would say.) Math is the shit. I always let my arts major friends know, once they get past the initial difficulty, math is fun, and they should give it a chance. Just doing my part to make everyone a little more employable.

[Ersa1a]

@barbarossaaaa Just curious, here. In your chemical engineering classes, do you use any textbooks by JD Seader? He's a friend of my family.

[didles123]

@MrFalcon675 He also could have used l'Hospital's rule as well because 0/0 is an indeterminate form. He skipped a step which was pretty self evident.