I'm pretty new to derivatives and I'm not sure if I understand the concept well.

Let's say we have function $f(x) = x^2$.If we set $x=3$, our example point is set at coordinates $(3,9)$.Using formula we can find derivative of $f(x)$: $f'(x)=6$.

My question is: what does this number $6$ mean? From what I've read, it should be slope of the chosen point. Is the slope defined by one number? What is actual use of such slope?



3 Answers 3


It's the slope of the line tangent to the graph of $f(x)=x^2$ at the point $(3,9)$.

enter image description here

Say you have a differentiable function (that is a function with a derivative) $f$. Now consider any point $(x,f(x))$ on the curve $f$. No matter how curvy the function is at that point, you can always zoom in enough so that the function looks pretty much like a straight line in a very small interval around that point. The derivative $f'(x)$ is the slope of the line most similar to the curve at that point.

enter image description here

Notice how in the left picture, if we zoom in enough the graph looks like a line. The derivative is the slope of that line. The right picture on the other hand doesn't look like a line when you zoom in really close. That tells us that that function doesn't have a derivative at the point right where it makes a 'v'.

  • $\begingroup$ Great explanation. +1 $\endgroup$
    – Ivo Terek
    Sep 26, 2015 at 19:02
  • $\begingroup$ Great explanation! If I could additionally ask you: let's say we found slope of the point.What is use of such slope? Why would anyone be interested in finding those 'lines' in curvy functions? $\endgroup$
    – Jay Sky
    Sep 26, 2015 at 19:04
  • $\begingroup$ @JaySky It tells you how fast the function is changing. For instance, say that $f(t)$ is a function that describes your position on a racetrack. The derivative at a point would tell you how fast your position was changing at that point -- i.e. your speed. $\endgroup$ Sep 26, 2015 at 19:05
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    $\begingroup$ @JaySky You'll have to take more caclulus and see! But here's a basic list: The derivative of position with respect to time is velocity. The derivative of velocity with respect to time is acceleration. The derivative of total cost with respect to the number of units is marginal cost. The derivative of potential energy with respect to position is force. The derivative of voltage with respect to position is the electric field. The derivative of population with respect to time is the population growth rate. I'm out of room, there's a million more. These all make more sense if you stick with it. $\endgroup$ Sep 26, 2015 at 19:08
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    $\begingroup$ In your example, say that $f(x)=x^2$ describes how far you've fallen, after falling off a cliff after a time of $x$ seconds (that's not really the equation which describes falling, but let's pretend). Then $f(3)=9$ tells you that after $3$ seconds you've fallen $9$ feet. $f'(3)=6$ tells you that after $3$ seconds your speed is $6$ feet per second. Does that make it more clear? $\endgroup$ Sep 26, 2015 at 19:12

If you have $f(x) = x^2$, then $f'(3) = 6$ is the slope of the tangent line to the graph of $f$ at the point $(3,f(3)) = (3,9)$.

In general, given a differentiable function $f: \Bbb R \to \Bbb R$, and $a \in \Bbb R$, $f'(a)$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.

Armed with this interpretation, try to convince yourself that $f(x) = x^2$ has $f'(0) = 0$ just by looking at the graph, for example.


Notice, for a given function $y=f(x)$ the derivative $f'(x)$ shows the slope of the tangent line to the curve $y=f(x)$ at a given point say $(3, 9)$.

The slope is very important to find out the equation of the tangent at a given point on the curve.


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