{"id":3708,"date":"2026-05-26T12:10:03","date_gmt":"2026-05-26T12:10:03","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=3708"},"modified":"2026-05-26T14:10:18","modified_gmt":"2026-05-26T14:10:18","slug":"derivative-of-a-function","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/derivative-of-a-function.html","title":{"rendered":"Derivative of a Function: Step-by-Step Explanation Using the Limit"},"content":{"rendered":"

The derivative of a function is one of the basic concepts of calculus<\/a>. This idea is where a deeper understanding of how a function changes begins. If the value of a function changes, a natural question arises: how fast does this change happen? The derivative gives the answer to this question.<\/p>\n

In this article, we will look at the definition of the derivative and find out how to find the derivative of a function without ready-made differentiation rules. In other words, we will work without derivative tables and use only the limit.<\/p>\n

Derivative of a Function: Definition Using the Limit<\/h2>\n

Let the function \\( f(x) \\) be defined in some neighborhood of the point \\( a \\). This means that the function must be defined not only at the point \\( a \\) itself, but also at points that are close enough to it.<\/p>\n

Now take another point near \\( a \\), namely \\( a+h \\), where \\( h\\neq 0 \\). Here, \\( h \\) is the increment of the argument. In simple words, we move a little away from the point \\( a \\) to the right or to the left.<\/p>\n

\"Illustration<\/p>\n

Then the value of the function at the initial point is \\( f(a) \\), and the value of the function at the new point is \\( f(a+h) \\). So, the change in the value of the function has the form \\( f(a+h)-f(a) \\).<\/p>\n

However, the difference between the function values alone does not yet show how fast the function changes. That is why this difference is compared with the increment of the argument \\( h \\). For this, we form the ratio:<\/p>\n

\\[
\n\\frac{f(a+h)-f(a)}{h}.
\n\\]<\/p>\n

This fraction shows the average rate of change of the function on the interval from \\( a \\) to \\( a+h \\). But we are not interested in the whole interval. We want to understand the behavior of the function near the point \\( a \\). What should we do for this? We need to bring the point \\( a+h \\) closer to the point \\( a \\), that is, make the increment \\( h \\) tend to zero.<\/p>\n

If the limit<\/p>\n

\\[
\n\\lim_{h\\to 0}
\n\\frac{f(a+h)-f(a)}{h},
\n\\]<\/p>\n

exists, then this limit is called the derivative of the function \\( f(x) \\) at the point \\( a \\).<\/p>\n

So, we write:<\/p>\n

\\[
\nf'(a)=
\n\\lim_{h\\to 0}
\n\\frac{f(a+h)-f(a)}{h}.
\n\\]<\/p>\n

Thus, the derivative of a function at the point \\( a \\) is the limit of the ratio of the change in the function value to the change in the argument, as the increment of the argument tends to zero.<\/p>\n

If such a limit exists and is finite, then the function is called differentiable at the point \\( a \\).<\/p>\n

Main Notations for the Derivative<\/h3>\n

The derivative of the function \\( y=f(x) \\) can be written in different ways:<\/p>\n

\\[
\nf'(x), \\qquad
\ny’, \\qquad
\n\\frac{dy}{dx}, \\qquad
\n\\frac{d}{dx}f(x).
\n\\]<\/p>\n

The notation \\( \\frac{d}{dx} \\) is read as \u201ctake the derivative of the function \\( f(x) \\) with respect to the variable \\( x \\)\u201d<\/em>. If we need to write the value of the derivative at the point \\( a \\), we can use, for example, the following notations:<\/p>\n

\\[
\nf'(a), \\qquad
\n\\left.\\frac{dy}{dx}\\right|_{x=a}.
\n\\]<\/p>\n

So, different notations may look different, but they are all connected with the same action \u2014 finding the derivative.<\/p>\n

Finding the Derivative: Sequence of Main Steps<\/h2>\n

Now let\u2019s move from the derivative at one point to the general formula. If, instead of a specific point \\( a \\), we take an arbitrary point \\( x \\), then we can look not only for the number \\( f'(a) \\), but for the derivative as a function \\( f'(x) \\).<\/p>\n

Let the function \\( y=f(x) \\) be given. By definition, its derivative has the form:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

So, to find the derivative by definition, we act step by step.<\/p>\n

First, we give the argument \\( x \\) an increment \\( h \\). Then, instead of \\( x \\), we need to substitute \\( x+h \\) into the formula of the function. In this way, we get the new value of the function \\( f(x+h) \\).<\/p>\n

Next, we find the difference between the new and the initial values of the function: \\( f(x+h)-f(x) \\).<\/p>\n

After that, we form the ratio of this difference to the increment of the argument:<\/p>\n

\\[
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

At this stage, it is very important to simplify the expression we get. Why is this necessary? Because often, after directly substituting \\( h=0 \\), we get the indeterminate form \\( \\frac{0}{0} \\).<\/p>\n

That is why we first need to expand the brackets, combine like terms, factor out a common factor, or cancel \\( h \\), if this is possible. Only after that do we move on to the limit.<\/p>\n

So, the last step is to find the limit:<\/p>\n

\\[
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

If this limit exists, then it is the derivative of the function:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

In a shortened form, the algorithm can be written like this:<\/p>\n

\\[
\nf(x)
\n\\rightarrow
\nf(x+h)
\n\\rightarrow
\nf(x+h)-f(x)
\n\\rightarrow
\n\\frac{f(x+h)-f(x)}{h}
\n\\rightarrow
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

So, finding the derivative by definition is a step-by-step transition from the change in the argument to the change in the function, and then to the limit of their ratio. This approach shows exactly why the derivative is connected with the instantaneous rate of change of a function.<\/p>\n

Derivative of a Function: Finding It by Definition in Practice<\/h2>\n

Now let\u2019s move on to practice. Examples help us see how the definition of the derivative works not only as a formula, but also as a clear step-by-step algorithm. We will move gradually: from simpler functions to those where a little more algebra is needed.<\/p>\n

Example 1. Find the derivative of the function \\( f(x)=3\\cdot x-2 \\)<\/h3>\n

By the definition of the derivative, we write:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

First, let\u2019s find \\( f(x+h) \\). To do this, we substitute \\( x+h \\) instead of \\( x \\) in the formula of the function:<\/p>\n

\\[
\nf(x+h)=3\\cdot (x+h)-2.
\n\\]<\/p>\n

Now expand the brackets:<\/p>\n

\\[
\nf(x+h)=3\\cdot x+3\\cdot h-2.
\n\\]<\/p>\n

Next, find the difference \\( f(x+h)-f(x) \\):<\/p>\n

\\[
\nf(x+h)-f(x)
\n=
\n(3\\cdot x+3\\cdot h-2)-(3\\cdot x-2).
\n\\]<\/p>\n

Expand the brackets and combine like terms:<\/p>\n

\\[
\nf(x+h)-f(x)
\n=
\n3\\cdot x+3\\cdot h-2-3\\cdot x+2=3\\cdot h.
\n\\]<\/p>\n

Now substitute the obtained expression into the derivative formula:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{3\\cdot h}{h}.
\n\\]<\/p>\n

Since \\( h\\neq 0 \\), we can cancel \\( h \\):<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}3=3.
\n\\]<\/p>\n

So, the derivative of the function \\( f(x)=3\\cdot x-2 \\) is equal to \\( 3 \\). This means that this linear function changes at a constant rate.<\/p>\n

Example 2. Find the derivative of the function \\( f(x)=x^2 \\)<\/h3>\n

By definition, we have:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

Find \\( f(x+h) \\):<\/p>\n

\\[
\nf(x+h)=(x+h)^2.
\n\\]<\/p>\n

Expand the square of the sum:<\/p>\n

\\[
\nf(x+h)=x^2+2\\cdot x\\cdot h+h^2.
\n\\]<\/p>\n

Now find the difference between the new and initial values of the function:<\/p>\n

\\[
\nf(x+h)-f(x)
\n=
\n(x^2+2\\cdot x\\cdot h+h^2)-x^2.
\n\\]<\/p>\n

After combining like terms, we get:<\/p>\n

\\[
\nf(x+h)-f(x)=2\\cdot x\\cdot h+h^2.
\n\\]<\/p>\n

Substitute this into the derivative formula:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{2\\cdot x\\cdot h+h^2}{h}.
\n\\]<\/p>\n

Factor \\( h \\) out of the numerator:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{h\\cdot (2\\cdot x+h)}{h}.
\n\\]<\/p>\n

Cancel \\( h \\):<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}(2\\cdot x+h).
\n\\]<\/p>\n

Now we pass to the limit. When \\( h\\to 0 \\), the expression \\( 2\\cdot x+h \\) tends to \\( 2\\cdot x \\). Therefore,<\/p>\n

\\[
\nf'(x)=2\\cdot x.
\n\\]<\/p>\n

Example 3. Find the derivative of the function \\( f(x)=x^3 \\)<\/h3>\n

We use the definition of the derivative:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

Find \\( f(x+h) \\):<\/p>\n

\\[
\nf(x+h)=(x+h)^3.
\n\\]<\/p>\n

Expand the cube of the sum:<\/p>\n

\\[
\nf(x+h)=x^3+3\\cdot x^2\\cdot h+3\\cdot x\\cdot h^2+h^3.
\n\\]<\/p>\n

Now find the difference:<\/p>\n

\\[
\nf(x+h)-f(x)
\n=
\n(x^3+3\\cdot x^2\\cdot h+3\\cdot x\\cdot h^2+h^3)-x^3.
\n\\]<\/p>\n

Cancel \\( x^3 \\) and \\( -x^3 \\):<\/p>\n

\\[
\nf(x+h)-f(x)=3\\cdot x^2\\cdot h+3\\cdot x\\cdot h^2+h^3.
\n\\]<\/p>\n

Substitute this into the derivative formula:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{3\\cdot x^2\\cdot h+3\\cdot x\\cdot h^2+h^3}{h}.
\n\\]<\/p>\n

Factor \\( h \\) out of the numerator:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{h\\cdot (3\\cdot x^2+3\\cdot x\\cdot h+h^2)}{h}.
\n\\]<\/p>\n

Cancel \\( h \\):<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n(3\\cdot x^2+3\\cdot x\\cdot h+h^2).
\n\\]<\/p>\n

When \\( h\\to 0 \\), the terms \\( 3\\cdot x\\cdot h \\) and \\( h^2 \\) tend to zero. Therefore,<\/p>\n

\\[
\nf'(x)=3\\cdot x^2.
\n\\]<\/p>\n

Example 4. Find the derivative of the function \\( f(x)=\\frac{1}{x} \\)<\/h3>\n

We consider the case \\( x\\neq 0 \\). We also need to remember that \\( x+h\\neq 0 \\), because the expression \\( f(x+h) \\) must be defined.<\/p>\n

By the definition of the derivative:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

Find \\( f(x+h) \\):<\/p>\n

\\[
\nf(x+h)=\\frac{1}{x+h}.
\n\\]<\/p>\n

Now substitute this into the formula:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{\\frac{1}{x+h}-\\frac{1}{x}}{h}.
\n\\]<\/p>\n

In the numerator, we have the difference of fractions. Let\u2019s bring them to a common denominator:<\/p>\n

\\[
\n\\frac{1}{x+h}-\\frac{1}{x}
\n=
\n\\frac{x-(x+h)}{x\\cdot (x+h)}.
\n\\]<\/p>\n

Simplify the numerator:<\/p>\n

\\[
\nx-(x+h)=x-x-h=-h.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\frac{1}{x+h}-\\frac{1}{x}
\n=
\n\\frac{-h}{x\\cdot (x+h)}.
\n\\]<\/p>\n

Now return to the derivative formula:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{\\frac{-h}{x\\cdot (x+h)}}{h}.
\n\\]<\/p>\n

Division by \\( h \\) can be replaced by multiplication by \\( \\frac{1}{h} \\):<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{-h}{x\\cdot (x+h)}\\cdot \\frac{1}{h}.
\n\\]<\/p>\n

Cancel \\( h \\):<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{-1}{x\\cdot (x+h)}.
\n\\]<\/p>\n

When \\( h\\to 0 \\), we have \\( x+h\\to x \\). Therefore,<\/p>\n

\\[
\nf'(x)=
\n-\\frac{1}{x^2},
\n\\qquad x\\neq 0.
\n\\]<\/p>\n

Example 5. Find the derivative of the function \\( f(x)=\\sqrt{x} \\)<\/h3>\n

For this function, we need to remember its domain: \\( x\\geq 0 \\). We will find the derivative by definition for \\( x>0 \\), so that the expressions under the square root are well-defined near the point \\( x \\).<\/p>\n

By definition:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{f(x+h)-f(x)}{h}.
\n\\]<\/p>\n

Find \\( f(x+h) \\):<\/p>\n

\\[
\nf(x+h)=\\sqrt{x+h}.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{\\sqrt{x+h}-\\sqrt{x}}{h}.
\n\\]<\/p>\n

If we substitute \\( h=0 \\) immediately, we get the indeterminate form \\( \\frac{0}{0} \\). So, we need to transform the expression. To do this, multiply the numerator and the denominator by the conjugate expression \\( \\sqrt{x+h}+\\sqrt{x} \\):<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{\\sqrt{x+h}-\\sqrt{x}}{h}
\n\\cdot
\n\\frac{\\sqrt{x+h}+\\sqrt{x}}{\\sqrt{x+h}+\\sqrt{x}}.
\n\\]<\/p>\n

In the numerator, we get the difference of squares:<\/p>\n

\\[
\n(\\sqrt{x+h}-\\sqrt{x})\\cdot (\\sqrt{x+h}+\\sqrt{x})
\n=
\n(x+h)-x.
\n\\]<\/p>\n

That is, the numerator is equal to \\( h \\). So, we have:<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{h}{h\\cdot (\\sqrt{x+h}+\\sqrt{x})}.
\n\\]<\/p>\n

Cancel \\( h \\):<\/p>\n

\\[
\nf'(x)=
\n\\lim_{h\\to 0}
\n\\frac{1}{\\sqrt{x+h}+\\sqrt{x}}.
\n\\]<\/p>\n

When \\( h\\to 0 \\), we have \\( \\sqrt{x+h}\\to \\sqrt{x} \\). Therefore,<\/p>\n

\\[
\nf'(x)=
\n\\frac{1}{2\\sqrt{x}},
\n\\qquad x>0.
\n\\]<\/p>\n

Note.<\/strong> Even simple examples show that finding the derivative directly by definition takes a lot of time and is often quite labor-intensive. That is why, in practice, derivatives are usually found using differentiation rules and formulas.<\/p><\/blockquote>\n

What to Read Next: Topics to Continue With<\/h2>\n

After the definition of the derivative, it is logical to move on to topics that show how this concept works in different problems. In this way, learning becomes step-by-step: first, the meaning of the derivative, then the rules, and after that, more complex cases of differentiation.<\/p>\n

    \n
  1. Mechanical and Geometric Meaning of the Derivative: How to Understand Its Meaning<\/a> \u2014 This article will explain how the derivative describes the speed of motion and the slope of the tangent line to the graph of a function at a given point.<\/li>\n
  2. Rules for Differentiating Functions and the Table of Derivatives: Calculations Without Limits<\/a> \u2014 This material will explain the main rules of differentiation and show how the table of derivatives makes solving typical problems easier.<\/li>\n
  3. Derivative of a Composite Function: Applying the Chain Rule<\/a> \u2014 This article will look at how to find the derivative of a composite function and how to correctly identify the outer and inner parts.<\/li>\n<\/ol>\n

    Derivative of a Function: Flowchart for Program Implementation<\/h2>\n

    If you are interested in programming, try going one step further and implementing the flowchart below in your favorite language: Pascal<\/em>, Python<\/em><\/a>, C++<\/em>, JavaScript<\/em>, or any other language. The idea is simple, but very useful.<\/p>\n

    The program calculates the derivative of the function \\( y=x^2+3\\cdot x-5 \\) at a given point in two ways: analytically, using the formula \\( f'(x)=2\\cdot x+3 \\), and approximately, using the definition \\( \\frac{f(x+h)-f(x)}{h} \\). Then it compares the results and shows the difference between them.<\/p>\n

    Isn\u2019t this an interesting way to see how a mathematical idea turns into a working algorithm?<\/p>\n

    \"Flowchart<\/p>\n","protected":false},"excerpt":{"rendered":"

    The derivative of a function is one of the basic concepts of calculus. This idea is where a deeper understanding<\/p>\n","protected":false},"author":1,"featured_media":3731,"comment_status":"open","ping_status":"closed","sticky":false,"template":"template-centered.php","format":"standard","meta":{"footnotes":""},"categories":[358],"tags":[539,537,540,363,538],"class_list":["post-3708","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-derivative-and-differential","tag-calculus","tag-derivative","tag-derivative-definition","tag-derivative-examples","tag-function-derivative"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3708","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/comments?post=3708"}],"version-history":[{"count":20,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3708\/revisions"}],"predecessor-version":[{"id":3711,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3708\/revisions\/3711"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media\/3731"}],"wp:attachment":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media?parent=3708"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/categories?post=3708"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/tags?post=3708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}