# In the definition of the derivative formula why is the limit of delta x approaching 0 when you are not suppose to substitute 0 for x?

The derivative of $f(x) = \sqrt{x}$ is $\frac{1}{2\sqrt{x}}$

The tangent slope is $f(x)$ is $\frac{1}{2}$ after the limit. We wait to put the limit $x \rightarrow 0$ until we find the tangent slope so why is it in the definition of a derivative?

its definition is:

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

Where $x = \delta x$

• In 2-d the slope of the tangent line and the derivative are the same. sqrt(x) has changing tangent line... it is not always 1/2. That is reflected in its derivative being a function of x. Let me know if you need more help understanding. – Matthew Levy Oct 26 '14 at 19:43
• The slope of the tangent line to $f(X)$ is not $\frac{1}{2}$ at every point $x$. we can check this by viewing the graph. – Eoin Oct 26 '14 at 19:56

So, a slope of a line that passes through two points can be described as "rise over run", "change in y over change in x."

It is with that thought in mind that the limit definition of derivative was formed. The following are equivalent.

$\lim\limits_{h\rightarrow 0} \frac{f(x+h) - f(x)}{x+h - x} = \lim\limits_{c\rightarrow x} \frac{f(c) - f(x)}{c - x} = \frac{\mathrm{d}}{\mathrm{d}x}[f(x)] = f'(x) = \frac{\mathrm{d}f}{\mathrm{d}x}(x)$

The idea is that for a curve which is potentially more confusing than a straight line, you might get an incorrect result if you looked at the "average rate of change" by picking two points to draw a line between far enough away, but as the two points picked get closer and closer together, it becomes more and more accurate.

In doing one of these calculations (for example, the first form), if you tried plugging in the value of $h=0$ you will get a result of $\frac{f(x) - f(x)}{x-x} = \frac{0}{0}$ which is multivalued and could equal literally anything. You must first do some clever algebra to get it to a point that there is no $\frac{0}{0}$ or other similar "indeterminate form." Since we are saying that $h$ is not technically a zero (but is a number as close to zero as we want), we can technically cancel h's on top and on bottom if we can separate it well enough.

As for the difference between slope and derivative, the derivative at x gives you the slope of the tangent line at x and will have the same value, however the derivative is a function and can change value depending on what $x$ is. We don't commonly define "slope" to be a function that changes, but rather a static property.

• I highly recommend watching Dr. Herb Gross's videos titled Calculus Revisited from MiT open courseware if you have the time: youtube.com/watch?v=rXOGLlKuvzU for part1 of lecture 1. Lecture 1 is a crash course on analytic geometry and lecture 2 defines the derivative. – JMoravitz Oct 26 '14 at 20:01

Well,I answered my own question -

X is a variable at this step in the process, we don't know its value and so we cannot solve for it. We first take the derivative, then we can find the slope tangent at any point in the graph.

To understand better;

The derivative The end result of a derivative of an equation allows us to substitute a point into x and the equation will equal the value of the slope tangent to that particular x-coordinate on the graph.

The Slope tangent at a point After we've taken a derivative of a function we substitute whatever value we want that is on the x-coordinate of that function and come up with the slope tangent at a point.