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# ​ Differentiation by first principle

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Nov 24, 2018

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The most important thing to implant into your brain is that the first derivative at a point IS the gradient of the tangent at that point.  They are the same thing!

Take a loook at this​

https://www.desmos.com/calculator/w6urx7lydx

It is easiest to do the second bit first.

Find the gradient of the line joining (2,5)  to  (3,22)  both of which lie on the graph  $$y=x^3-2x+1$$

$$gradient = \frac{22-5}{3-2}=\frac{17}{1}=17$$

if (5,2) is fixed but (3,22) can vary i could say

$$\text{gradient of secant}= \frac{f(x)-5}{x-2} \qquad \text{where in this case x=3}$$

Now look at my desmos graph again:

It is not explained as well as it should be becasue it is not the gradient that should be moving and thereby making the second x intercept change, it should be the other way around but that is harder to draw.

If you play with the graph you can see that as the gradient greduces from 17 the second x value will moves closer to 2.

so

$$\text{gradient of tangent at (2,5)}\\=\displaystyle\lim_{x\rightarrow 2}\frac{f(x)-5}{x-2}\\ =\displaystyle\lim_{x\rightarrow 2}\frac{x^3-2x+1-5}{x-2}\\ =\displaystyle\lim_{x\rightarrow 2}\frac{x^3-2x-5}{x-2}\\ =\displaystyle\lim_{x\rightarrow 2}\frac{(x-2)(x^2+2x+2)}{x-2}\\ =\displaystyle\lim_{x\rightarrow 2}(x^2+2x+2)\\ =4+4+2\\ =10$$

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Nov 24, 2018