# Understanding the Definition of a derivative as slope of a tangent line

I'm trying to understand the derivative and am wondering why the derivative is described as the slope of the tangent line and not the slope of a function itself.

Say $f(x) = 2x+5$ where $\frac{d}{dx}=2$, which is the slope of $2x+5$

We would say in algebra that $2$ is the slope of $f(x)$ but in calculus we now say that it is the slope of the tangent line.

Is it incorrect to say that the derivative is the slope of the function at an infinitesimally small change in $x$?

• The intuition is that the derivative is the slope at an infinitesimally small region around $x$ is correct but we don't say slope as the only functions with a slope are flat. For example if $f(x)=x^2$ then it makes sense to ask what the slope is of the tangent at a point but if you say "What is the slope of this function?" the answer is "Well it depends on where you look at it". What this boils down to is that slope is defined only for linear equations where as derivatives are defined for a bunch more. May 27, 2015 at 16:19
• The slope of a function is not defined; the slope of a line is.
– zhw.
May 27, 2015 at 17:52

But here there is a problem: what we means exactly for a "straight line tangent to a point"? We know that a straight line is well defined when we have two points, so, given a point $(x,f(x))$ and a point $(x+h,f(x+h))$ we can write the slope of the function that passes between these points as: $$\frac{f(x+h)-f(x)}{h}$$ than we define the tangent as the line that passes through $(x,f(x))$ and has slope the limit of this quotient when $h \rightarrow 0$ if this limit exists.
• The problem I face is when it comes to actual functions. The derivative of $x^2$ is $2x$. Now this derivative is not tangent to the said function; also the derivative seems stationary. For this case, how does the definition make sense? What about $x^3$? Apr 6, 2020 at 14:34