I have a doubt. Please correct me if I’m mistaken. Derivative is considered as rate of change = slope isn’t it?
If the derivative of $x^2$ is $2x$, that means slope = $2x$.
But $f(2)=4$, so the slope should be ${4\over 2}=2$; but here $2x=4$.
Please help me regarding this.
Original text:
Please help me regarding this doubt. I've just started my derivative lessons
Please find this image :
Thanks
$\endgroup$ 82 Answers
$\begingroup$Rather you should think of it as a function that can give you a slope.
For example, in your case of $$x^2$$
The derivative is $2x$.
What that means is that at any x-coordinate of $x^2$, you can get the slope, by plugging in that x coordinate, into your derivative.
It's like slope, but not the exact same.
Rather it's a function that gives you slope.
Consider $$\sin(x)$$
It's derivative is $\cos(x)$. $\cos(x)$ is not a slope. That doesn't make sense. Rather its a function that can give you a slope of the original function. So if we wanted the slope at $x=0$ of $\sin(x)$, we plug in $x=0$ into it's derivative and we see the slope is $\cos(0)=1$. Remember, slope is a number, not a function.
$\endgroup$ $\begingroup$You write that the slope of $f(x)=x^2$ at $x=2$ is ${f(2)\over 2}$. This is incorrect; the value of ${f(x)\over x}$ has nothing to do with the slope.
Here's a concrete way to see that $f(x)\over x$ can't have anything to do with the slope. Consider the function $g(x)=x^2+100$. If you draw this, it's clear that it is "parallel" to $f(x)=x^2$. So the slope of $g$ at $x=2$ should be the same as the slope of $f$ at $x=2$. But $g(2)\over 2$ is $52$, whereas $f(2)\over 2$ is $2$. So these numbers can't be the slopes!
I think you're misremembering the usual rise-over-run definition of the slope of a line: the slope of the line through $(x_1, y_1)$ and $(x_2, y_2)$ is $${y_2-y_1\over x_2-x_1}.$$ But note that this is very different from e.g. ${y_2\over x_2}$ - it takes into account two points!
EDIT: Bobbie's comment below suggests that you might be looking at the line from $(0, 0^2)$ to $(2, 2^2)$. This line indeed has slope $2$; however, it's not the tangent line! If you draw it, you'll notice that - where it meets the parabola - the parabola is much steeper (= has greater slope). So - if this is what you're doing - you have measured the slope of a line, just not the right line.
$\endgroup$ 4
