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Confusion on the given tutorial.


Vastor
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Hey guys,

 

based on 2nd post in the link

 

[math]\frac{(1+h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = \frac{h(2 + h)}{h} = 2 + h[/math]

we can cut and all of that if x = 1.

but how about if we express the answer in terms of x?

[math]\frac{(x+h)^2 - x}{h} = \frac{x^2 + 2xh + 2h - x}{h} = ?[/math]

seems like quadratic equation, doesn't it will produce 2 answer? (where we only get one gradient(tangent) only, right?)

btw, it's still not available to be answered because the h can't be 0.

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Hey guys,

 

now, I'm having problem get some intuitive on Chain Rule.

 

Chain Rule Proof

 

 

where it said:

 

[math] \frac{g(x + h) - g(x)}{h} - g'(x) = v \rightarrow 0 [/math], as [math] h \rightarrow 0 [/math]

 

but then, when I tried to:

 

[math] \frac{g(x + 0) - g(x)}{0} - g'(x) = \frac{g(x) - g(x)}{0} - 1 = \frac{0}{0} - 1 = 0? [/math]

 

doesn't [math] \frac{0}{0} [/math] = undefined?

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  • 3 weeks later...

You are taking a limit so you can avoid [math] \frac{0}{0} [/math].

 

Remember [math] h \rightarrow 0 [/math] means that [math] h [/math] is approaching (or tends to) [math] 0 [/math] and should not be simply substituted in.

 

 

 

What you must remember is that you are trying to find the gradient of a tangent slope at a single point.

The general idea behind what you are doing is: You need to make a secant line between two different points, if they were the same point you could not find the slope as you would have [math] \frac{0}{0} [/math]. This however does not give us an accurate tangent slope. To overcome this problem you let the distance between those two points approach 0 (so the slope of the secant line gets ever closer to the tangent slope you are looking for), and that is where taking the limit helps us.

 

I suggest, as did the tree; go back and learn limits fully before trying to tackle the chain rule.

Edited by Pixel
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