Please forgive me if it is dumb question as I am not very good with math. My question is, I am a little confused as to why we associate derivative with the slope of tangent line. Can any one please elaborate ?

Regards, Ahsan

  • $\begingroup$ The derivative of a function $f(x)$, typically denoted by $f'(x)=\frac{df}{dx}$, describes a slope at any given $x$ value. For example, if one were to plug in, say $x=2$, then $f'(2)$ is the instantaneous slope of $f(x)$ at $x=2$. Hope this clarifies a little. :) $\endgroup$
    – Cookie
    Mar 19, 2014 at 6:27
  • 2
    $\begingroup$ Think about how we compute the slope of a line (i.e. "rise over run" or $\Delta y/ \Delta x$). Now look at the definition of the derivative and see if it looks anything like that for the points $(x, f(x))$ and $(x+h, f(x+h))$. $\endgroup$
    – foobar1209
    Mar 19, 2014 at 6:28
  • $\begingroup$ Does the animated graph here en.wikipedia.org/wiki/Tangent#Tangent_line_to_a_curve help at all? $\endgroup$ Mar 19, 2014 at 6:30
  • $\begingroup$ Before you learn the concept of the derivative you do NOT know what the tangent line really is! In order to define the tangent, you already have to know what the derivative is...A statement like 'the derrivative is the slope of the tangent line' is very confusing: what is meant is 'the tangent is the line which slope is the derivative at this point' $\endgroup$
    – Blah
    Mar 19, 2014 at 6:40
  • $\begingroup$ math.stackexchange.com/questions/1053482/… $\endgroup$
    – user301988
    May 28, 2017 at 23:20

2 Answers 2


enter image description here

Let there be some change in $x$ by $h$. If we were to draw a line cutting through the initial and resultant point, then we would have a line passing through $(x, f(x))$ and $(x+h, f(x+h))$ as shown in the diagram above. (This line is the secant line, as some may recognize).

The derivative at a point is essentially the resultant change in $f(x)$ as a ratio of an infinitesimally small change in $x$. This means that the derivative at a point is $\frac{f(x+h) - f(x)}{(x+h)-x}$, for some infinitesimally small $h$. But this is exactly the gradient of the secant line shown above.

If we were to make $h$ very small to produce the "infinitesimally small change in $x$" (i.e. let $h \to 0$), then notice that the secant line above eventually becomes the tangent line at the point $(x, f(x))$. The derivative at that point then eventually becomes equal to the slope of the tangent line.

  • $\begingroup$ damn you beat me to it. :p +1 $\endgroup$
    – Guy
    Mar 19, 2014 at 6:37

Let's imagine we have a curve. Now we take any two points and take a line joining these two points. This line is called a secant as it cuts the curve at at least two points( there may be more but that is none of our concern). Now if you imagine taking these points closer to each other, you will see the secant becomes close to the tangent line. If you keep decreasing the distance between these points, and make them arbitrarily close, the secant line therefore becomes the tangent. This needs to be rigorously proved, but it is a good way to get a "feel" of what is going on, which I believe is what you are asking for.

Now the slope($m$) of this secant line should be equal to the slope of the tangent.

Thus $$m = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}$$

Taking $x_2=x_1+h$ and taking the limit $h\to0$


This limit is called the "derivative" and is equal to the slope as we wanted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.