I am just learning derivatives and I found the derivative of $4x-x^2$ to be $4-2x$. At point $(1,3)$ the tangent line is $2x+1$. Now when I graph this, the derivative $4-2x$ cuts through the function $4x-x^2$. Does that mean the derivative is the secant line?
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$\begingroup$The answer generally speaking is no. The derivative of a function at a point always represents the slope of the tangent line at that point. Sometimes the tangent line (by coincidence) crosses the curve somewhere else technically making it also a secant line, but there is no special meaning in this.
Based on your comment I misunderstood your qustion. The answer is still no in general. Try plotting the derivative of y=x^3.
$\endgroup$ 16 $\begingroup$There is a strange misunderstanding that the graph of $f'$ should be related to the graph of a tangent line of $f$. Even more when the graph of $f'$ is itself a line.
But given any number $a$, the value of $f'(a)$ keeps track of the magnitude of a slope, namely, it is the value we assign to $m$ in $$\tau(x)=m(x-a)+f(a)$$ so that $$\frac{f(x)-\tau(x)}{x-a}\to 0 \text{ when } x\to a$$
Without appealing technicalities, this means $f'(a)$ is the slope of the line that best approximates $(\to 0)\;\;$ $f$ in the vicinity of $a\;$ ($x\to a$). When we plot $f'$, we're seeing how the value of the slope at $x$ varies.
ADD As an example, the derivative of $3x+x^3$ is a second degree polynomial $3+3x^2$. This is not a line at all: it simply tells us that the slope of $3x+x^3$ is always positive, and that it gets steeper and steeper for both large positive values and large negative values of $x$. It also says $3x+x^3$ has no horizontal tangents, since the derivative is never zero.
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