Let $f:A \rightarrow \mathbb R$ be a differentiable function, where $A$ is an open interval of real numbers.

Let $a \in A$.

Then the tangent of $f$ at $a$ is defined by the equation $f(a)+f'(a)(x-a)$.

I understand the intuition behind choosing a point $f(a)$ through which the tangent will definitely pass through and another point $f'(a)$ - which, since it defines the slope of $f$ at $a$, the tangent line will surely pass through too.

Now, I don't understand the intuition behind why $f'(a)$ is multiplied by $(x-a)$ and not just $x$ - if we multiply $f'(a)$ by every real number, in a space with Cartesian coordinates we are going to have the range of the function be every point in the line which passes through $f(a)$ and $f'(a)$.

Why don't we define the tangent of $f$ at $a$ to be simply $f(a)+f'(a)x$?

  • $\begingroup$ Let $g(x)$ be the function of the tangent line at $x=a$. One must have $g(a)=f(a)$. $\endgroup$ – Jack Mar 1 '17 at 21:31
  • 2
    $\begingroup$ Consider the case $f(x)=x+1$. What do you think the tangent line at $x=1$ should be? $\endgroup$ – Jack Mar 1 '17 at 21:33

Because it doesn't work that way.

To start with, $f(a)$ is not a point. It's only a number. An example of a point on the graph of $f$ is $(a, f(a))$--two coordinates, $a$ and $f(a),$ to plot a point in the plane.

Next, $f'(a)$ is the slope of the tangent line. You may have realized that already, but remember that the equation of a line in the $x,y$ plane satisfies $y = b + mx$ if and only if $m$ is the slope and $b$ is the $y$-intercept of the line. So if we can figure out a number $b$ such that the tangent line at $(a, f(a))$ goes through the point $(0,b),$ then we can write $y = b + (f'(a))x.$

There is a more generally-applicable form of the equation of (almost) any line in the plane, which lets you write the equation for a line easily once you know its slope and any point that it passes through (not necessarily where it intersects the $y$ axis). For a line with slope $m$ passing through the point $(x_1,y_1),$ the equation is $$ y = y_1 + m(x - x_1). \tag1 $$ The equation can also be written in a slightly more symmetric form, $$ y - y_1 = m(x - x_1), $$ which tells us this is the equation of the line we get if we translate the line $y=mx$ upward $y_1$ units and $x_1$ units to the right so that it goes through the point $(x_1,y_1).$

For the tangent line, instead of an arbitrary point named $(x_1,y_1)$ we have one named $(a,f(a)),$ and instead of an arbitrary slope $m$ we have slope $f'(a).$ So where do these new symbols fit into Equation $1,$ and what does it look like after we update it to describe the tangent line?

Of course you still can write an equation for the tangent line in intercept-slope form if you really want to. Just observe that the equation of the tangent line says $$ y = f(a)+f'(a)(x-a) = \left( f(a) - af'(a) \right) + (f'(a))x, $$ so the $y$-intercept of the tangent line is $f(a) - af'(a).$

  • $\begingroup$ A fundamental remark is that the equation of the tangent $y=f(a)+f'(a)(x-a)$ is the beginning ot Taylor expansion of $f$ around pooint a, i.e., $f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2+...$. In other words, it is the best affine approximation of function $f$ , up to 2nd degree terms... $\endgroup$ – Jean Marie Mar 1 '17 at 22:53
  • $\begingroup$ @JeanMarie That's very true, but I designed this answer to be read by someone who is just starting to learn about derivatives and their relationship to the tangent lines of graphs of functions. I don't expect the reader to know about Taylor expansions yet. ... Maybe I can add something at the end about "things you will see later." $\endgroup$ – David K Mar 1 '17 at 23:38
  • $\begingroup$ @I agree with you : I appreciate your understanding of the level at which the answers should be given. $\endgroup$ – Jean Marie Mar 2 '17 at 7:07

What is the equation of the line passing through $(a,b)$ of slope $m?$ It is $y = b + m(x-a).$ So then, what is the equation of the line passing through $(a,f(a))$ of slope $f'(a)?$ It is

$$y = f(a) + f'(a)(x-a).$$


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.