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September 26th, 2014, 02:07 PM   #2
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 Originally Posted by DarkX132 he wrote down the equation of the graph and then he did something completely unexpected: y=x^2+3x-3 or,y=2x+3
He probably meant $\dfrac{dy}{dx}=2x+3$, as you write later. The expression $\dfrac{dy}{dx}$ is the derivative, and you can read more about it in Wikipedia (see especially the "Derivatives of elementary functions" section). It is also the slope of the tangent line at point $(x,y(x))$. September 26th, 2014, 03:23 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra Could your teacher have written $$y^\prime = 2x + 3$$ This is another way of writing $$\frac{\mathbb d y}{\mathbb d x} = 2x + 3$$ As to understanding it, I assume you haven't met the theory of limits yet, so the best bet will be the way Isaac Newton developed it. We can approximate the gradient of a curve $y=f(x)$ by taking any two points $(x,f(x))$ and $(x+c,f(x+c))$ on the curve. We can then join the two points with a straight line. The gradient of the curve (at $x$) is approximately the gradient of the line. And the gradient of the line is$$\frac{\text{change in y}}{\text{change in x}} = \frac{f(x+c)-f(x)}{(x+c) - x} = \frac{f(x+c)-f(x)}{c}$$ Now, for large $c$, this approximation is poor, but the smaller we make $c$, the better the approximation becomes. So let's make $c$ really, really, small. As small as you can imagine. Clearly, the approximation is now really, really good. So let's see what our approximation gives $$\frac{f(x+c)-f(x)}{c} = \frac{\left( (x+c)^2 + 3(x+c) - 3\right) - \left( x^2 +3x -3\right)}{c} = \frac{\cancel{x^2} + 2xc + c^2 + \cancel{3x} + 3c - \cancel3 - \cancel{x^2} - \cancel{3x} + \cancel3}{c} = 2x + 3 + c$$ And now comes Newton's fudge. He says that because $c$ is really, really small, we can ignore it. So the gradient of the curve at any point $x$ is$$2x+3$$ In the years since Newton developed this method, it has been refined and made rigorous, but we still use his 'difference quotient' as the starting point. Thanks from greg1313 September 26th, 2014, 10:15 PM #4 Newbie   Joined: Sep 2014 From: Bangladesh Posts: 4 Thanks: 0 Oh I get it now Thanks Tags algebra, equation, graphical, jump ,

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# in a graph how do you draw 2 cm to 2units y axis

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