Can you deduce ##\tan(\theta) = \frac {df} {dx}## from this graph?

In summary, by drawing a vertical line through a point on a graph and considering the triangle formed by that point, the intersection of the vertical line with the x-axis, and the intersection of the tangent with the x-axis, we can deduce that the tangent of an angle is equal to the derivative of the function at that point. This is because a tangent has the same derivative as the function at the point where it touches the function. Additionally, the slope of the tangent line is equal to the tangent of the angle formed by the dotted line and the x-axis. This is applicable even if the line is not a radius of a circle.
  • #1
I_laff
41
2
Could someone explain to me how from this graph you can deduce that ##\tan(\theta) = \frac {df} {dx}##.
Thanks

upload_2018-6-11_11-55-30.png
 

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  • #2
Draw a vertical line going through point P and consider the triangle made by P, the intersection of the vertical line with the x-axis and the intersection of the tangent with the x axis. Then take into account that a tangent has the same derivative as the function at the point where it touches the function (by definition).
 
  • #3
I think it's by definition that CP and CQ are the same, line p denotes the circle, which has rise/slide df/dx as sin/cos which is tangent of the angle at P, which is the same angle as the dotted line makes with the x axis. If line p wasn't a radius, the slope would not be tan.
 

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