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I know that the purpose of differentiation is to find the rate of change of a function. But i don't really understand how this is concept can be used to find the tangent of a specific point. It would be greatly appreciated if you can explain this to me.

Gonçalo
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user377786
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  • Do you know the definition of derivative of a function $f$ at a point $x_0$? If so, that is the limit of the slope of a secant line joining the points $(x_0,f(x_0))$ and $(x_0+h,f(x_0+h))$ when $h$ tends to $0$. But then the secant line through the above two points tends to the tangent line at $(x_0,f(x_0))$. – Gibbs Nov 03 '23 at 17:02
  • See https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/acf917af6fb88659fe64f36e2b7c5d57_MIT18_01SCF10_Ses1b.pdf – Vasili Nov 03 '23 at 17:17
  • Welcome to MSE! <> One question to ask is, "What by definition is a tangent line?" If the definition is the line whose slope at $(x_0, f(x_0))$ is $f'(x_0)$, which is not unreasonable, there's not much to say. ;) – Andrew D. Hwang Nov 03 '23 at 21:06

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On one hand, differentiation has been indeed conceived as a measure of the variation of a given function $f(x)$, when on the other hand its geometric interpretation corresponds to the slope of the tangent line of the graph of the same function. Actually, the link between these two points of view lies in the fact that the tangent slope is precisely the rate of change of $f(x)$, because the higher the slope (in absolute value), the bigger the variation of $f(x)$. This link is even "materialized" algebraically through the (first-order) Taylor expansion of $f(x)$, since $f(x) = f(a) + f'(a)(x-a) + \mathcal{O}(x^2)$ coincides with the tangent line of $f(x)$ at $x = a$.

Abezhiko
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