In school I was taught to memorize this formula.
$ \frac {d} {dx} (\cos x) = - \sin x $

However, recently I found out a proof using the first principle (under the "Derivatives" chapter), but could not understand the proof.

The formula for first principle is as follows:
$\ f'(x) = \lim_{h\to 0} \frac {f(x+h)-f(x)} {h}$

As per the proof, I need to substitute the $\cos x$ in the above equation as shown below and arrive the result.
$\ f'(x) = \lim_{h\to0} \frac {\cos(x+h)-\cos x} {h}$

I understand the math behind this and I get how the result is $-\sin x$.

How does this work? What is that $h$ and why it is tending to $0$.

  • 2
    $\begingroup$ This is the definition of the derivative... $\endgroup$ – YoTengoUnLCD Feb 6 '16 at 18:15
  • $\begingroup$ I can list the entire proof here. But this "first principle" is not convincing enough. When I saw how the first principle formula was arrived at, it showed a graph with a curve and a tangent next to it. Now, how can I relate the two ? $\endgroup$ – Prasanna Feb 6 '16 at 18:18
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    $\begingroup$ The first principle computation of the derivatives of sine and cosine typically uses addition theorems for these trigonometric functions and somewhere deep inside requires the computation of one single specific limit: $\lim_{x\to 0}\frac{\sin x}{x}=1$. One has to be careful not to fall into circular argumentation at that point! A geometrically inclined proof of that limit typically compares the areas of certain triangles and circular sectors to establish $\sin x<x<\tan x$ for $0<x<\frac\pi2$ or the like. $\endgroup$ – Hagen von Eitzen Feb 6 '16 at 18:25
  • $\begingroup$ @hagenconeitzen Yes, the geometric argument works well heuristically. More rigorously, one can use the integral definition of the arcsine and quite easily prove the limit. Obviously, this requires an understanding of Riemann integration and knowledge regarding inverse functions and their derivatives. Alternatively, one can use the power series representation for the sine and cosine, but this requires knowledge of convergence theorems for differentiating power series. So, I agree with you that the geometric reasoning makes sense at the intro level. $\endgroup$ – Mark Viola Feb 6 '16 at 18:58
  • $\begingroup$ You need some "manipulations" : $\cos(x+h)=\cos x \cos h - \sin x \sin h$. $\endgroup$ – Mauro ALLEGRANZA Jun 14 '17 at 15:18

The derivative is just the slope of the line tangent to a point on the curve. This leads us to the natural way to arrive at the definition you wrote by writing what the slope of the line through two points on the curve is. Then by taking the limit as these two points get very close together (as the difference $h$ goes to $0$), you see that the slope of this line approaches the slope of the tangent at a point.
See the figure below:
enter image description here


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