Vastor Posted April 29, 2012 Share Posted April 29, 2012 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. Link to comment Share on other sites More sharing options...

the tree Posted April 29, 2012 Share Posted April 29, 2012 (edited) You just made a tiny mistake, you should be looking at: [math]\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}[/math] you'll find it works out easily. Edited April 29, 2012 by the tree 1 Link to comment Share on other sites More sharing options...

Vastor Posted May 7, 2012 Author Share Posted May 7, 2012 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? Link to comment Share on other sites More sharing options...

the tree Posted May 7, 2012 Share Posted May 7, 2012 Taking the limit as h -> 0 is different to just substituting in h=0. You'll need a better idea of limits before trying to work your way through that proof. Link to comment Share on other sites More sharing options...

Pixel Posted May 29, 2012 Share Posted May 29, 2012 (edited) 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 May 29, 2012 by Pixel Link to comment Share on other sites More sharing options...

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