# How can derivatives represent tangents?

I am taking a Introduction to Calculus course and am struggling to understand how derivatives can represent tangent lines.

I learned that derivatives are the rate of change of a function but they can also represent the slope of the tangent to a point. I also learned that a derivative will always be an order lower that the original function.

For example: $$f(x) = x^3 and f'(x) = 3x^2$$

What I fail to understand is that how can $$3x^2$$ represent the slope of the tangent line if it is not a linear function?

Wouldn't this example mean that the slope or the tangent itself is a parabola?

• If you have a fixed real number $x_0$, then the slope at that point is $3x_0^2$ in your example. Since this works for every real $x_0$, we get a function again. Feb 20, 2020 at 10:19
• Notice that slope is a real number, so it can be a function! The slope is not the same as the tangent line. If the tangent line is something like $y=kx+c$, then $k$ is the slope, so surely it can be any number. Calculus tells us that $k=f'(x)$. Feb 20, 2020 at 10:22
• $f'(x)$ is used to calculate the slope. It isn't the equation of the tangent. At $x = a$, the equation of the tangent is $y = f(a)+f'(a)(x-a)$ (this is simply point-slope form). For instance, $f'(2) = 12$, meaning that the slope of the tangent at $x = 2$ is $12$. Of course, using the point $(2, 8)$ means you can write the equation for the tangent line too: $y-8 = 12(x-2)$. Feb 20, 2020 at 10:26
• @KM101 I don't understand your equation. If $x=a$ then the tangent line would be $f(a)$ itself. Wouldn't it? Feb 20, 2020 at 10:34
• The equation for any line given its slope $m$ and a point $\left(x_0, y_0\right)$ is $y-y_0 = m\left(x-x_0\right)$. At $x = a$, $y = f(a)$ and $m = f'(a)$ (the slope of the tangent), so that's where the equation comes from. I'm not sure what you mean by $f(a)$ though. Could you explain what you mean by that? Feb 20, 2020 at 10:37

The derivative represents the slope of the tangent, not the equation of a tangent line.

For understanding why it is so, we delve into the question of 'what is the derivative?', the fundamental idea of finding the derivative is taking a point on the curve and another point, which is extremely close to it, and computing the slope of the line through those two points. This is reflected in the definition of the derivative, which I assume you are familiar with.

$$\lim_{h \to 0} \frac{ f(x+h)-f(x)}{h}$$

if you look at any curve, you would notice that that a line tangent to the curve at one point won't be tangent it to at another by the very definition of the tangent. And hence it is understandable that the derivative of a function is in fact another function which relates x-coordinate of a point on a curve to the slope of the line tangent to it.

Finally, if you really wanted, you could find the equation of tangent as well. For this, you have to simply use the 'point slope formula' of the line

$$\frac{y-y_o}{x-x_o} = {\frac{dy}{dx}}\biggr\rvert_{x_o}$$

where the slope is the derivative evaluated at x-coordinate of the point where the tangent meets the curve.

• why is the mathjax not working??? Feb 20, 2020 at 13:22
• I find your explanation about the derivative being a function to be very good and intuitive. Feb 20, 2020 at 22:42

What happens is that, for each $$a$$ in the domain of $$f$$, $$f'(a)$$ is the slope of the the tangent to the graph of $$f$$ at the point $$\bigl(a,f(a)\bigr)$$.

So, if $$f(x)=x^3$$, since $$f'(x)=3x^2$$, the slope of the tangent to the graph of $$f$$ at the point $$(1,1)$$ is $$3$$, and therefore that tangent line is the line $$3(x-1)+1\bigl(=3x-2\bigr)$$.

• The formula defining the derivative function is not itself the equation of the tangent; this formula gives you , for each tangent ( one tangent for each point $$(x, f(x))$$ of the graph of $$f$$ ), the slope of this line. And a slope is a number.

The main point here is that the derivative function is a function that sends back numbers as outputs ( not lines, not tangents). One and only one number far every permissible input $$x$$.

• To understand this, remember that for all point $$(x, f(x))$$ of the graph (such that there is a tangent to the graph at this point), this tangent will have the form :

$$y = mx + b$$.

The number $$m$$ is the slope of the tangent. You may think of it as a percent ( in the same way as we ordinarily think of the slope of a road in terms of %).

For example, the slope of the line $$y = 0,5x +2$$ has slope $$0,5$$, that is, $$50$$%. The slope of the line $$6x + 10$$ has slope $$6$$, that is $$600$$%. The slope of $$y=0x+5=5$$ is $$O$$ ( = $$0$$%). The slope of $$y= -2x +40$$ is $$-2$$ = $$- 200$$% ( These are arbitrary examples, not related to the $$x^3$$ function).

• So, for each input $$x$$, the derivative gives as output the number $$m$$ ( that is, the slope) of the tangent to the graph at point $$( x, f(x))$$.

• The beauty is that, although the tangents will ( ordinarily) have various slopes, although the outputs of function $$f'(x)$$ will be different for different $$x$$ values ( inputs), we are often able to find a rule defining a constant numerical relation between the value of $$x$$ and the corresponding slope. For example, for $$f(x)=x²$$, it can be proved that $$f'(x)$$ ( the slope of the tangent to the graph of $$f$$ at $$(x, f(x))$$) is always the double of x ! This is what means the differentiation rule : $$\frac {d} {dx}x^2$$ $$=$$ $$2\times x$$.

Note : this number that is sent back as output is formally defined as a limit, namely, the limit , as $$h$$ approaches $$0$$ , of the ratio

$$\frac {f(x+h) - f(x)} { (x+h) - x}$$ = $$\frac{change- in-y}{change-in-x}$$

This shows that the slope of the tangent happens to be identical to the instantaneous rate of growth of the original function $$f$$ at the point $$( x, f(x))$$. This is why, in fact, we are interested in these slopes.

Note : you can use the number $$f'(a)$$ to find the equation of the tangent at a given point $$( a, f(a))$$. Since $$f'(a)$$ is " the $$m$$ ( = slope) of this tangent" ,the equation of this line will have the form : $$y = f'(a)x + b$$. The fact that you also know one point of this tangent , namely, the point $$(a, f(a))$$, allows you ( with some algebra) to recover the number $$b$$ , and finally the whole equation of the tangent at this point $$( a, f(a))$$.

• Examples with $$f(x)= x^3$$ and consequently $$f'(x)= 3x^2$$:

For $$x= 1$$ , the slope is $$f'(1)$$ = $$3\times1^2$$= $$3$$ = $$300$$%

So, at $$( 1, f(1))$$ , the slope of the tangent to the graph of $$f$$ is $$300$$%. Quite a big slope.

For $$x= 2$$ , the slope is $$f'(2)$$=$$3\times2^2$$= 12 = $$1200$$%

So, at $$( 3, f(3))$$ , the slope of the tangent to the graph of $$f$$ is $$1200$$%. A huge slope!

• I'd avoid the term 'percent', since it implies the value of the derivative is out of $100$. There's also a few minor things you could possibly word better. "except in cases where $f(x)$ is linear" - what about constant functions? ""growth" here can be positive or negative" - What about $f'(x)=0$? Also, your attached image says "For some functions, there are points... ...of the graph such that the graph has no tangent line..." which would seem to imply that points where $f'$ does not exist must be isolated, which is untrue.
– Jam
Mar 7, 2020 at 21:07
• My answer aims mainlly at correcting the " category mistakes" that prevent the OP from understanding the derivative function concept. I think that the main point he missed is that the derivative function (1) is a function and (2) sends back numbers (3) that each one of these numbers is ( or can be viewed as) the slope of the tangent at $( x, f(x))$. This is why I used the " percent" phrasing, since a slope is a rate. I think this is intuitive and helps to take a good start , for we usually use the percent phrasing when talking about slopes of roads, for example. ( It's the case in France).
– user655689
Mar 7, 2020 at 21:16
• The danger of the percent phrasing, however, is that we naturally tend to think that a percent is a number between 0 and 1 ( = 100%). But slopes (1) can be negative (2) and can be greater than 1, that is, more than 100%.
– user655689
Mar 7, 2020 at 21:28