So, when differentiating, it is worth working step by step. First, we determine what we have in front of us: a sum, a difference, a product, a quotient, a composite function, or an expression with a constant. Then we apply the appropriate rule. And only after that do we use the necessary formula from the table of derivatives.<\/p>\n
Also, let\u2019s not forget about preliminary transformations. Sometimes an expression can be simplified even before differentiation. For example, a fraction can be reduced, powers can be combined, and parentheses can be expanded. Isn\u2019t it easier to make the expression simpler first and only then find the derivative? This approach often helps avoid unnecessary calculations and mistakes.<\/p>\n
Differentiation Rules for Functions and the Table of Derivatives: Practical Use of Formulas<\/h2>\n
Now let\u2019s move on to practice. After all, differentiation eules become truly clear when we apply them to specific examples. We will work step by step: first, we determine the structure of the function, then choose the rule, and after that use the table of derivatives.<\/p>\n
Example 1. Find the derivative of the function \\( f(x)=x^5-3\\cdot x^2+7\\cdot x-10 \\)<\/h3>\n
We have an algebraic sum of several terms. So, first, we apply the rule for the derivative of a sum and difference. This means that each term can be differentiated separately:<\/p>\n
\\[
\nf'(x)=(x^5)’-(3\\cdot x^2)’+(7\\cdot x)’-(10)’.
\n\\]<\/p>\n
Now let\u2019s use the table of derivatives. For a power function, we have the formula \\( (x^n)’=n\\cdot x^{n-1} \\). We also remember that the derivative of a constant is equal to zero.<\/p>\n
Then:<\/p>\n
\\[
\n\\begin{gathered}
\n(x^5)’=5\\cdot x^4,
\n\\\\[4pt]
\n(3\\cdot x^2)’=3\\cdot (x^2)’=3\\cdot 2\\cdot x=6\\cdot x,
\n\\\\[4pt]
\n(7\\cdot x)’=7,
\n\\\\[4pt]
\n(10)’=0.
\n\\end{gathered}
\n\\]<\/p>\n
Therefore,<\/p>\n
\\[
\nf'(x)=5\\cdot x^4-6\\cdot x+7.
\n\\]<\/p>\n
Example 2. Find the derivative of the function \\( f(x)=(x^2+1)\\cdot \\sin(x) \\)<\/h3>\n
Here the function is a product of two expressions. The first factor is \\( x^2+1 \\), and the second factor is \\( \\sin(x) \\). So, we need to use the product rule:<\/p>\n
\\[
\n(u(x)\\cdot v(x))’=u'(x)\\cdot v(x)+u(x)\\cdot v'(x).
\n\\]<\/p>\n
Let \\( u(x)=x^2+1 \\), and \\( v(x)=\\sin(x) \\). Then:<\/p>\n
\\[
\n\\begin{gathered}
\nu'(x)=(x^2+1)’=2\\cdot x,
\n\\\\[4pt]
\nv'(x)=(\\sin(x))’=\\cos(x).
\n\\end{gathered}
\n\\]<\/p>\n
Now substitute these derivatives into the product rule:<\/p>\n
\\[
\nf'(x)=2\\cdot x\\cdot \\sin(x)+(x^2+1)\\cdot \\cos(x).
\n\\]<\/p>\n
Therefore, the derivative is:<\/p>\n
\\[
\nf'(x)=2\\cdot x\\cdot \\sin(x)+(x^2+1)\\cdot \\cos(x).
\n\\]<\/p>\n
Example 3. Find the derivative of the function \\( f(x)=\\frac{x^3+2\\cdot x}{x^2+1} \\)<\/h3>\n
We have a quotient of two functions. So, we use the quotient rule:<\/p>\n
\\[
\n\\left(\\frac{u(x)}{v(x)}\\right)’=
\n\\frac{u'(x)\\cdot v(x)-u(x)\\cdot v'(x)}{\\bigl(v(x)\\bigr)^2}.
\n\\]<\/p>\n
Let the numerator be \\( u(x) \\), and the denominator be \\( v(x) \\):<\/p>\n
\\[
\n\\begin{gathered}
\nu(x)=x^3+2\\cdot x,
\n\\\\[4pt]
\nv(x)=x^2+1.
\n\\end{gathered}
\n\\]<\/p>\n
Now let\u2019s find their derivatives. For the numerator, we get:<\/p>\n
\\[
\nu'(x)=(x^3+2\\cdot x)’=3\\cdot x^2+2.
\n\\]<\/p>\n
For the denominator:<\/p>\n
\\[
\nv'(x)=(x^2+1)’=2\\cdot x.
\n\\]<\/p>\n
Substitute everything into the quotient rule:<\/p>\n
\\[
\nf'(x)=
\n\\frac{(3\\cdot x^2+2)\\cdot (x^2+1)-(x^3+2\\cdot x)\\cdot 2\\cdot x}{(x^2+1)^2}.
\n\\]<\/p>\n
Next, simplify the numerator. First, expand the first product:<\/p>\n
\\[
\n(3\\cdot x^2+2)\\cdot (x^2+1)=3\\cdot x^4+3\\cdot x^2+2\\cdot x^2+2=3\\cdot x^4+5\\cdot x^2+2.
\n\\]<\/p>\n
Now expand the second product:<\/p>\n
\\[
\n(x^3+2\\cdot x)\\cdot 2\\cdot x=2\\cdot x^4+4\\cdot x^2.
\n\\]<\/p>\n
Therefore, the numerator is equal to:<\/p>\n
\\[
\n3\\cdot x^4+5\\cdot x^2+2-(2\\cdot x^4+4\\cdot x^2)=x^4+x^2+2.
\n\\]<\/p>\n
So, finally, we have:<\/p>\n
\\[
\nf'(x)=\\frac{x^4+x^2+2}{(x^2+1)^2}.
\n\\]<\/p>\n
Example 4. Find the derivative of the function \\( f(x)=\\ln(x^2+1) \\)<\/h3>\n
We have a composite function. The outer function is the natural logarithm, and the inner function is \\( x^2+1 \\). So, here we need to use the chain rule:<\/p>\n
\\[
\n(f(g(x)))’=f'(g(x))\\cdot g'(x).
\n\\]<\/p>\n
Let the inner function be:<\/p>\n
\\[
\ng(x)=x^2+1.
\n\\]<\/p>\n
Then the given function can be viewed as \\( \\ln(g(x)) \\). From the table of derivatives, we know that \\( (\\ln(x))’=\\frac{1}{x} \\). But since under the logarithm we have not just \\( x \\), but the expression \\( x^2+1 \\), we also need to multiply by the derivative of the inner function.<\/p>\n
Find the derivative of the inner function:<\/p>\n
\\[
\ng'(x)=(x^2+1)’=2\\cdot x.
\n\\]<\/p>\n
Now apply the chain rule:<\/p>\n
\\[
\nf'(x)=\\frac{1}{x^2+1}\\cdot 2\\cdot x.
\n\\]<\/p>\n
Therefore,<\/p>\n
\\[
\nf'(x)=\\frac{2\\cdot x}{x^2+1}.
\n\\]<\/p>\n
Example 5. Find the derivative of the function \\( f(x)=\\frac{x^2-1}{x} \\)<\/h3>\n
At first glance, we have a quotient. So, we could immediately apply the quotient rule. But is this always the most convenient way? No. Sometimes it is better to simplify the function before differentiating.<\/p>\n
Divide each term in the numerator by \\( x \\):<\/p>\n
\\[
\n\\frac{x^2-1}{x}=\\frac{x^2}{x}-\\frac{1}{x}.
\n\\]<\/p>\n
Therefore,<\/p>\n
\\[
\nf(x)=x-\\frac{1}{x}.
\n\\]<\/p>\n
Now the function has become much simpler. Let\u2019s apply the rule for the derivative of a difference:<\/p>\n
\\[
\nf'(x)=(x)’-\\left(\\frac{1}{x}\\right)’.
\n\\]<\/p>\n
According to the table of derivatives, \\( (x)’=1 \\), and \\( \\left(\\frac{1}{x}\\right)’=-\\frac{1}{x^2} \\). Then:<\/p>\n
\\[
\nf'(x)=1-\\left(-\\frac{1}{x^2}\\right).
\n\\]<\/p>\n
Therefore,<\/p>\n
\\[
\nf'(x)=1+\\frac{1}{x^2}.
\n\\]<\/p>\n
The Next Step in Differentiation: Topics for Further Study<\/h2>\n
After studying differentiation rules and the table of derivatives, it is worth moving on to topics where the derivative is found in a less standard way. This is exactly where we can clearly see how basic formulas work in more complex situations. So, what should we consider next?<\/p>\n
- \n
- Derivative of an Implicitly Defined Function: Differentiation Without an Explicit Expression<\/a> \u2014 This article will explain how to find a derivative when a function is given by an equation, and the required variable is not expressed explicitly.<\/li>\n
- Derivative of a Parametrically Defined Function: Working with an Auxiliary Parameter<\/a> \u2014 This topic will explain how to find a derivative when the coordinates depend not on each other, but on a common parameter.<\/li>\n
- Logarithmic Differentiation: A Convenient Method for Complex Expressions<\/a> \u2014 This article will show how logarithms help simplify the differentiation of products, quotients, and power expressions.<\/li>\n<\/ol>\n
Differentiation Rules and Table of Derivatives: An Algorithm for Your Own Code<\/h2>\n
If you are interested in programming, try turning your knowledge of derivatives into a small practical project. The flowchart below shows an algorithm that finds the value of the derivative of a polynomial at a given point: the program takes the degree, the coefficients, and the value of \\( x \\), and then calculates the result.<\/p>\n
- Derivative of a Parametrically Defined Function: Working with an Auxiliary Parameter<\/a> \u2014 This topic will explain how to find a derivative when the coordinates depend not on each other, but on a common parameter.<\/li>\n
![\"Flowchart](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/06\/differentiation-rules-and-table-of-derivatives2.jpg\")
<\/p>\n","protected":false},"excerpt":{"rendered":"