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# ​ Differentiation by first principle

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1 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.

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