# Relation between the tangent to a curve and the first derivative of a function

There is a relation between the tangent to a curve of a function and the first derivative of that function. However, how do I show that connection? How can you explain it to someone so that it becomes clear and understandable?

• You should check out the wikipedia entry for derivative. en.wikipedia.org/wiki/Derivative. Figures 1-3 are particularly illuminating. – jspecter Jun 14 '11 at 17:06
• Perhaps this might help. Especially from the paragraph that begins "Now, does the line joining $A$ and $B$ have a slope that really approaches the slope of the tangent?" – Arturo Magidin Jun 14 '11 at 20:46

My answer is going to be very informal but geared toward gaining some intuition. I'm assuming "nice curves" and drawing lots of pictures:

Remember that the derivative at $x$ is essentially defined as the limit of the slopes of secant lines near $x$.

This should provide some understanding of the relation you're talking about.
Draw a line through $f(x)$ and $f(x_1)$ where $x_1$ is "close to" $x$.
Then draw a line through $f(x)$ and $f(x_2)$ where $x_2$ is halfway between $x$ and $x_1$.
Then do this with $x_3$ halfway between $x$ and $x_2$, and so on.
Since you're drawing lines that intersect the curve at points $x$ and $x_n$, but $x_n$ keeps getting closer to $x$, "eventually" the line only intersects the curve at $x$.

Alternatively, notice that if you draw a line through $f(x)$ and a nearby $f(x')$, then via some sort of intermediate value argument, if the curve didn't have constant slope between those points, it must have gone "up" or "down" relative to your line between those points, and your line must not really be approximating the local slope of the curve.

Assume your curve $\gamma$ is given by $y=f(x)$ with $f(0)=0$ and that $\gamma$ is "curved upwards" at $(0,0)$. A tangent to $\gamma$ at $(0,0)$ is a line $y=m x$ through $(0,0)$ which has this point in common with $\gamma$ but stays below $\gamma$ otherwise. This means that $f(x)\geq m x$ for all $x$ near $0$. This implies $${f(x)-f(0)\over x-0}={f(x)\over x}\geq m \qquad(x>0)$$ and $${f(x)-f(0)\over x-0}={f(x)\over x}\leq m \qquad(x<0)\ .$$ If $f$ is differentiable at $0$ then both left sides of these equations have the same limit $m'$ when $x\to 0+$ resp. $x\to 0-$. The only possible common value is $m$ which is then at the same time the slope of the tangent to $\gamma$ and the value of $f'(0)$.

Suppose $f(x)=x^2$, so that $f(5)=25$, and that means the point $(5,25)$ is on the graph of the function. Then $f'(x) = 2x$, and so $f'(5) = 10$. The slope of the tangent line at the point where $x=5$ is therefore $10$. So the tangent line to the graph at that point is a line that passes through the point $(5,25)$ and whose slope is $10$.

That's a typical instance.