Gradient of a curve at a point

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apple2357

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Suppose i want to find the gradient of the curve y= x^2 at the point x=3
The standard approach is to differentiate and sub x=3 so the answer is 6
Why does this work though?
Consider (x-3)^2
Expanding and you get x^2-6x+9
The gradient is the negative of the x coefficient ( i.e. 6)
I have tried this with higher powers like x^3 etc and it works too? Still scratching my head over this...
 
I think you're saying that the derivative of x^2 at x=3 is the negative of the coefficient of the x term of the expansion of (x-3)^2.

To see why, try finding the derivative of x^2 at x=3 from the definition. What do you see?
 
The definition of the derivative; what other definition would I refer to?

But, yes, people do often seem to call the use of this definition "working from first principles". Give it a try.
 
Ok I thought you meant that the gradient of curve is defined as the gradient of the tangent to the curve at that point. But ok I will have a play with first principles.

The only hunch I had was something to do with repeated roots. But not sure that goes anywhere.
 
That idea goes back to Fermat who used it to find slopes of tangent line before Newton and Leibniz developed Calculus. If a line, y= mx+ b, cuts a curve, y= f(x) in two places then mx+ b= f(x) has two roots. If it is a tangent line at x= a then mx+ b= f(x) has x= a as a double root. In the case of \(\displaystyle f(x)= x^2\) that becomes \(\displaystyle mx+ b= x^2\) or \(\displaystyle x^2- mx- b= 0\). That will have, say, x= 3 as a double root if and only of \(\displaystyle x^2- mx- b= (x- 3)^2= x^2- 6x+ 9\). By inspection, m= 6 and b= -9. The tangent line to \(\displaystyle y= x^2\) at x= 3 is y= 6x- 9 and the gradient there is 6.
 
Ah very nice! Makes a lot of sense.
So you can use this approach to find gradients without differentiation!!
 
Sometimes; it doesn't work for all functions.

Have you tried finding the slope at x=3 "from first principles"? Here it is, starting from the definition of the derivative:

[MATH]f(x) = x^2[/MATH]​

[MATH]f'(x)=\lim_{x\to 3}\frac{f(3+h)-f(3)}{h}=\lim_{x\to 3}\frac{(h+3)^2-(3)^2}{h}=\lim_{x\to 3}\frac{h^2+6h+9-9}{h}=\lim_{x\to 3}\frac{h^2+6h}{h}=\lim_{x\to 3}(h+6)=6[/MATH]​

Do you see that the 6 is exactly the coefficient of x (here, h) in the expansion? That will be true for the limit anywhere.

Of course, that's true for this particular function, and extends to any power function [MATH]f(x) = x^n[/MATH]. It isn't true of functions that can't be expanded like that -- but it turns out that there is a sense in which any function (with some restrictions) can be expanded (into a typically infinite series) and the concept is still true.
 
Yes i did do it using first principles and could see that 6 reappeared in the expansion so that made sense too.
Thanks
 
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