Definition of the Derivative
Solved Problems
Example 9.
Using the limit definition find the derivative of the function \(f\left( x \right) = \frac{1}{{x - 1}}.\)
Solution.
Write the increment of the function:
\[\require{cancel}\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = \frac{1}{{x + \Delta x - 1}} - \frac{1}{{x - 1}} = \frac{{x - 1 - \left( {x + \Delta x - 1} \right)}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}} = \frac{{\cancel{\color{blue}{x}} - \cancel{\color{red}{1}} - \cancel{\color{blue}{x}} - \Delta x + \cancel{\color{red}{1}}}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}} = - \frac{{\Delta x}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}.\]
The difference quotient is given by
\[\frac{{\Delta y}}{{\Delta x}} = - \frac{{\frac{{\Delta x}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}}}{{\Delta x}} = - \frac{\cancel{\Delta x}}{{\cancel{\Delta x}\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}} = - \frac{1}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}.\]
Now we can pass to the limit:
\[f^\prime\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \left( { - \frac{1}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}} \right) = - \frac{1}{{\left( {x - 1} \right)\left( {x - 1} \right)}} = - \frac{1}{{{{\left( {x - 1} \right)}^2}}}.\]
Example 10.
Find the derivative of the function \(f\left( x \right) = \frac{2}{{{x^2}}}\) using the limit definition.
Solution.
First we write the function difference:
\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = \frac{2}{{{{\left( {x + \Delta x} \right)}^2}}} - \frac{2}{{{x^2}}} = \frac{2}{{{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}}} - \frac{2}{{{x^2}}} = \frac{{2{x^2} - 2\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)}}{{{x^2}\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)}} = \frac{{\cancel{2{x^2}} - \cancel{2{x^2}} - 4x\Delta x - 2{{\left( {\Delta x} \right)}^2}}}{{{x^2}\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)}} = - \frac{{4x\Delta x + 2{{\left( {\Delta x} \right)}^2}}}{{{x^2}\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)}}.\]
The difference quotient has the form
\[\frac{{\Delta y}}{{\Delta x}} = - \frac{{\frac{{4x\Delta x + 2{{\left( {\Delta x} \right)}^2}}}{{{x^2}\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)}}}}{{\Delta x}} = - \frac{{\left( {4x + 2\Delta x} \right)\cancel{\Delta x}}}{{{x^2}\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)\cancel{\Delta x}}} = - \frac{{4x + 2\Delta x}}{{{x^2}\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)}}.\]
Determine the derivative by calculating the limit of the difference quotient as \(\Delta x \to 0:\)
\[f^\prime\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = - \lim\limits_{\Delta x \to 0} \frac{{4x + 2\Delta x}}{{{x^2}\left( {{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)}} = - \frac{{4x}}{{{x^2} \cdot {x^2}}} = - \frac{4}{{{x^3}}}.\]
Example 11.
Find the derivative of the function \(y = \sqrt x .\)
Solution.
By the definition of derivative,
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{y\left( {x + \Delta x} \right) - y\left( x \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\sqrt {x + \Delta x} - \sqrt x }}{{\Delta x}}.\]
We now multiply numerator and denominator of this quotient by \({\sqrt {x + \Delta x} + \sqrt x }.\) Notice that
\[\left( {\sqrt {x + \Delta x} - \sqrt x } \right) \left( {\sqrt {x + \Delta x} + \sqrt x } \right) = {\left( {\sqrt {x + \Delta x} } \right)^2} - {\left( {\sqrt x } \right)^2} = \cancel{x} + \Delta x - \cancel{x} = \Delta x.\]
Then
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\sqrt {x + \Delta x} - \sqrt x }}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{\cancel{\Delta x}}{{\cancel{\Delta x}\left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} = \lim\limits_{\Delta x \to 0} \frac{1}{{\sqrt {x + \Delta x} + \sqrt x }} = \frac{1}{{2\sqrt x }}.\]
Example 12.
Determine the derivative of the cube root function \(f\left( x \right) = \sqrt[3]{x}\) using the limit definition.
Solution.
First we write the function difference:
\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = \sqrt[3]{{x + \Delta x}} - \sqrt[3]{x}.\]
The difference ratio has the form
\[\frac{{\Delta y}}{{\Delta x}} = \frac{{\sqrt[3]{{x + \Delta x}} - \sqrt[3]{x}}}{{\Delta x}}.\]
We multiply the numerator and denominator by the conjugate expression
\[{\left( {\sqrt[3]{{x + \Delta x}}} \right)^2} + \left( {\sqrt[3]{{x + \Delta x}}} \right)\sqrt[3]{x} + {\left( {\sqrt[3]{x}} \right)^2}.\]
After some algebra we get
\[\frac{{\Delta y}}{{\Delta x}} = \frac{1}{{{{\left( {\sqrt[3]{{x + \Delta x}}} \right)}^2} + \left( {\sqrt[3]{{x + \Delta x}}} \right)\sqrt[3]{x} + {{\left( {\sqrt[3]{x}} \right)}^2}}}.\]
Calculate the limit of the difference ratio as \(\Delta x \to 0:\)
\[f^\prime\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{1}{{{{\left( {\sqrt[3]{x}} \right)}^2} + {{\left( {\sqrt[3]{x}} \right)}^2} + {{\left( {\sqrt[3]{x}} \right)}^2}}} = \frac{1}{{3\sqrt[3]{{{x^2}}}}}.\]
Example 13.
Calculate the derivative of the cubic function \(y = {x^3}.\)
Solution.
If we change the variable \(x\) by an amount \(\Delta x\) the function receives the following increment:
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = {\left( {x + \Delta x} \right)^3} - {x^3}.\]
Expand the cube of the sum and simplify:
\[\Delta y = {\left( {x + \Delta x} \right)^3} - {x^3} = \cancel{x^3} + 3{x^2}\Delta x + 3x{\left( {\Delta x} \right)^2} + {\left( {\Delta x} \right)^3} - \cancel{x^3} = \left( {3{x^2} + 3x\Delta x + {{\left( {\Delta x} \right)}^2}} \right)\Delta x .\]
Hence, the derivative of a cubic function has the following form:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \left( {3{x^2} + 3x\Delta x + {{\left( {\Delta x} \right)}^2}} \right) = 3{x^2}.\]
Here we took into account that the terms \(3x\Delta x\) and \({{\left( {\Delta x} \right)}^2}\) approach zero as \(\Delta x \to 0.\)
Example 14.
Differentiate the power function \(f\left( x \right) = {x^4}\) using the limit definition.
Solution.
The increment of the function is written as
\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = {\left( {x + \Delta x} \right)^4} - {x^4}.\]
We use the product formula
\[{\left( {a + b} \right)^4} = {a^4} + 4{a^3}b + 6{a^2}{b^2} + 4a{b^3} + {b^4}.\]
Hence
\[\Delta y = \cancel{x^4} + 4{x^3}\Delta x + 6{x^2}{\left( {\Delta x} \right)^2} + 4x{\left( {\Delta x} \right)^3} + {\left( {\Delta x} \right)^4} - \cancel{x^4} = 4{x^3}\Delta x + 6{x^2}{\left( {\Delta x} \right)^2} + 4x{\left( {\Delta x} \right)^3} + {\left( {\Delta x} \right)^4}.\]
Then the difference quotient is given by
\[\frac{{\Delta y}}{{\Delta x}} = 4{x^3} + 6{x^2}\Delta x + 4x{\left( {\Delta x} \right)^2} + {\left( {\Delta x} \right)^3}.\]
It is clear that the derivative has the form
\[f^\prime\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = 4{x^3}.\]
Example 15.
Find the derivative of the sine function \(y = \sin x.\)
Solution.
Using the definition of derivative, we obtain
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\sin \left( {x + \Delta x} \right) - \sin x}}{{\Delta x}}.\]
Apply the trigonometric identity
\[\sin \alpha - \sin \beta = 2\sin \frac{{\alpha - \beta }}{2}\cos \frac{{\alpha + \beta }}{2}.\]
Then we have
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\sin \left( {x + \Delta x} \right) - \sin x}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{2\sin \frac{{\cancel{x} + \Delta x - \cancel{x}}}{2}\cos \frac{{x + \Delta x + x}}{2}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{2\sin \frac{{\Delta x}}{2}\cos \left( {x + \frac{{\Delta x}}{2}} \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{2\sin \frac{{\Delta x}}{2}}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \cos \left( {x + \frac{{\Delta x}}{2}} \right).\]
The first limit in this expression is
\[\lim\limits_{\Delta x \to 0} \frac{{2\sin \frac{{\Delta x}}{2}}}{{\Delta x}} = \lim\limits_{\frac{{\Delta x}}{2}\to 0} \frac{{\sin \frac{{\Delta x}}{2}}}{{\frac{{\Delta x}}{2}}} = 1.\]
Since \(\lim\limits_{\Delta x \to 0} \cos \left( {x + \frac{{\Delta x}}{2}} \right) = \cos x\), the derivative of the sine function is given by
\[y'\left( x \right) = {\left( {\sin x} \right)^\prime } = \cos x.\]
Example 16.
Find the derivative of the cosine function \(y = \cos x.\)
Solution.
Following the definition of derivative, we write it as the limit:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{y\left( {x + \Delta x} \right) - y\left( x \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\cos \left( {x + \Delta x} \right) - \cos x}}{{\Delta x}}.\]
Transform the difference of cosines to product by the formula
\[\cos\alpha - \cos\beta = - 2\sin\frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}.\]
As a result we have
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\cos \left( {x + \Delta x} \right) - \cos x}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\left( { - 2\sin\left( {x + \frac{{\Delta x}}{2}} \right)\sin\frac{{\Delta x}}{2}} \right)}}{{\Delta x}} = - \lim\limits_{\Delta x \to 0} \frac{{2\sin\frac{{\Delta x}}{2}}}{{\Delta x}} \cdot {\lim\limits_{\Delta x \to 0} \sin\left( {x + \frac{{\Delta x}}{2}} \right).}\]
The first and second limits in this expression have the following values:
\[\lim\limits_{\Delta x \to 0} \frac{{2\sin\frac{{\Delta x}}{2}}}{{\Delta x}} = \lim\limits_{\frac{{\Delta x}}{2} \to 0} \frac{{\sin\frac{{\Delta x}}{2}}}{{\frac{{\Delta x}}{2}}} = 1,\;\;\lim\limits_{\Delta x \to 0} \sin \left( {x + \frac{{\Delta x}}{2}} \right) = \sin x.\]
Hence, the derivative of the cosine function is
\[y'\left( x \right) = {\left( {\cos x} \right)^\prime } = - \sin x.\]
Example 17.
Find the derivative of the trigonometric function \(f\left( x \right) = \sin 2x\) using the limit definition.
Solution.
Write the increment of the function:
\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = \sin \left( {2\left( {x + \Delta x} \right)} \right) - \sin 2x = \sin \left( {2x + 2\Delta x} \right) - \sin 2x.\]
Use the trigonometric identity
\[\sin \alpha - \sin \beta = 2\sin \frac{{\alpha - \beta }}{2}\cos \frac{{\alpha + \beta }}{2}.\]
Hence
\[\Delta y = 2\sin \frac{{2x + 2\Delta x - 2x}}{2} \cos \frac{{2x + 2\Delta x + 2x}}{2} = 2\sin \Delta x\cos \left( {2x + \Delta x} \right).\]
The ratio \(\frac{{\Delta y}}{{\Delta x}}\) is given by
\[\frac{{\Delta y}}{{\Delta x}} = \frac{{2\sin \Delta x\cos \left( {2x + \Delta x} \right)}}{{\Delta x}}.\]
Calculate the limit of the ratio as \(\Delta x \to 0:\)
\[f^\prime\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{2\sin \Delta x\cos \left( {2x + \Delta x} \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{2\sin \Delta x}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \cos \left( {2x + \Delta x} \right).\]
The first limit is
\[\lim\limits_{\Delta x \to 0} \frac{{2\sin \Delta x}}{{\Delta x}} = 2\lim\limits_{\Delta x \to 0} \frac{{\sin \Delta x}}{{\Delta x}} = 2 \cdot 1 = 2.\]
The second limit is
\[\lim\limits_{\Delta x \to 0} \cos \left( {2x + \Delta x} \right) = \cos 2x.\]
Hence, the derivative is
\[f^\prime\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = 2 \cdot \cos 2x = 2\cos 2x.\]
Example 18.
Find an expression for the derivative of the exponential function \(y = {e^x}\) using the definition of derivative.
Solution.
The initial steps are standard: first we write the increment of the function \(\Delta y\) corresponding to the increment of the independent variable \(\Delta x:\)
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = {e^{x + \Delta x}} - {e^x} = {e^x}{e^{\Delta x}} - {e^x} = {e^x}\left( {{e^{\Delta x}} - 1} \right).\]
The derivative can be computed as the limit of the ratio of the increments:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{{e^x}\left( {{e^{\Delta x}} - 1} \right)}}{{\Delta x}}.\]
Since the function \(y = {e^x}\) in the numerator does not depend on \(\Delta x\), it can be taken out of the limit. Then the derivative takes the following form:
\[y'\left( x \right) = \left( {{e^x}} \right)^\prime = {e^x}\lim\limits_{\Delta x \to 0} \frac{{{e^{\Delta x}} - 1}}{{\Delta x}}.\]
We denote the resulting limit by \(L\) and calculate it separately. Notice that \({e^0} = 1\) and therefore we can write
\[L = \lim\limits_{\Delta x \to 0} \frac{{{e^{\Delta x}} - 1}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{{e^{\Delta x}} - {e^0}}}{{\Delta x}} = e'\left( 0 \right),\]
that is, this limit is the value of the derivative of the exponential function at zero. Hence,
\[y'\left( x \right) = {\left( {{e^x}} \right)^\prime } = {e^x}e'\left( 0 \right).\]
We have a relationship in which the required derivative is expressed in terms of the function \(y = {e^x}\) and its derivative at \(x = 0\). We prove that
\[L = e'\left( 0 \right) = 1.\]
For this we recall that the number \(e\) is defined as the infinite limit
\[e = {e^1} = \lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n},\]
and the number \(e\) to the power of \(\Delta x\) is, respectively, given by
\[e^{\Delta x} = \lim\limits_{n \to \infty } \left( {1 + \frac{{\Delta x}}{n}} \right)^n.\]
Next we apply the famous Newton's Binomial Theorem and expand the expression under the limit sign into binomial series:
\[\left( {1 + \frac{{\Delta x}}{n}} \right)^n = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n\\k \end{array}} \right){{\left( {\frac{{\Delta x}}{n}} \right)}^k}} .\]
Here \({\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right)}\) denotes the number of \(k\)-combinations of \(n\) elements. Return back to our limit \(L,\) which can now be written in the form:
\[L = \lim\limits_{\Delta x \to 0} \frac{{{e^{\Delta x}} - 1}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\lim\limits_{n \to \infty } \left[ {\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n\\k \end{array}} \right){{\left( {\frac{{\Delta x}}{n}} \right)}^k}} } \right] - 1}}{{\Delta x}}.\]
It is convenient to separate the first two terms in the binomial series: for \(k = 0\) and \(k = 1\). As a result, we obtain
\[L = \lim\limits_{\Delta x \to 0} \frac{{\lim\limits_{n \to \infty } \left[ {\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}
n\\
k
\end{array}} \right){{\left( {\frac{{\Delta x}}{n}} \right)}^k}} } \right] – 1}}{{\Delta x}}
= \lim\limits_{\Delta x \to 0} \frac{{\Delta x + \lim\limits_{n \to \infty } \sum\limits_{k = 2}^n {\left( {\begin{array}{*{20}{c}}
n\\
k
\end{array}} \right){{\left( {\frac{{\Delta x}}{n}} \right)}^k}} }}{{\Delta x}}
= {1 \text{ +}\;\;}\kern-0.3pt{ \lim\limits_{n \to \infty } \left[ {\lim\limits_{\Delta x \to 0} \left( {\sum\limits_{k = 2}^n {\left( {\begin{array}{*{20}{c}}
n\\
k
\end{array}} \right)\frac{{{{\left( {\Delta x} \right)}^{k – 1}}}}{{{n^k}}}} } \right)} \right]}.\]
Obviously, the sum of the series approaches zero as \(\Delta x \to 0\). Therefore, \(L = 1\). This means that the derivative of the exponential function \(y = {e^x}\) is the function itself:
\[y'\left( x \right) = {\left( {{e^x}} \right)^\prime } = {e^x}.\]
Example 19.
Find the derivative of the power function \(y = {x^n}.\)
Solution.
Following the general scheme, we write the increment of the function:
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = {\left( {x + \Delta x} \right)^n} - {x^n}.\]
In the last expression, we expand the sum to the power of \(n\) by the binomial formula, which has the following form:
\[ {\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right){a^{n - k}}{b^k}} = \left( {\begin{array}{*{20}{c}} n\\ 0 \end{array}} \right){a^n} + \left( {\begin{array}{*{20}{c}} n\\ 1 \end{array}} \right){a^{n - 1}}b + \left( {\begin{array}{*{20}{c}} n\\ 2 \end{array}} \right){a^{n - 2}}{b^2} + \ldots + \left( {\begin{array}{*{20}{c}} n\\ {n - 1} \end{array}} \right)a{b^{n - 1}} + \left( {\begin{array}{*{20}{c}} n\\ n \end{array}} \right){b^n} = {a^n} + n{a^{n - 1}} + \frac{{n\left( {n - 1} \right)}}{{2!}}{a^{n - 2}}{b^2} + \ldots + na{b^{n - 1}} + {b^n}. \]
Then the increment \(\Delta y\) is given by
\[\Delta y = {\left( {x + \Delta x} \right)^n} - {x^n} = n{x^{n - 1}}\Delta x + \frac{{n\left( {n - 1} \right)}}{{2!}}{x^{n - 2}}{\left( {\Delta x} \right)^2} + \ldots + nx{\left( {\Delta x} \right)^{n - 1}} + {\left( {\Delta x} \right)^n}.\]
We turn to the ratio of the increments and the corresponding limit:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \Big[ {n{x^{n - 1}}} + {\frac{{n\left( {n - 1} \right)}}{{2!}}{x^{n - 2}}\left( {\Delta x} \right) + \ldots + nx{{\left( {\Delta x} \right)}^{n - 2}} + {{\left( {\Delta x} \right)}^{n - 1}}} \Big].\]
This shows that all the terms except the first vanish as \(\Delta x \to 0.\) Therefore, the derivative of the power function is
\[y'\left( x \right) = {\left( {{x^n}} \right)^\prime } = n{x^{n - 1}}.\]
Example 20.
Find the derivative of the natural logarithm \(y = \ln x.\)
Solution.
We derive the derivative of the natural logarithm based only on the definition and without using any other differentiation rules. We write the expression for the derivative as the limit:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{y\left( {x + \Delta x} \right) - y\left( x \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\ln \left( {x + \Delta x} \right) - \ln x}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\ln \left( {\frac{{x + \Delta x}}{x}} \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \left[ {\frac{1}{{\Delta x}}\ln \left( {1 + \frac{{\Delta x}}{x}} \right)} \right].\]
Introduce the new variable \(n = \left| {\frac{x}{{\Delta x}}} \right|\). Obviously that \(n \to \infty\) as \(\Delta x \to 0\). In terms of the new variable \(n,\) the derivative and the limit are written as
\[y'\left( x \right) = \lim\limits_{n \to \infty } \left[ {\frac{n}{x}\ln \left( {1 + \frac{1}{n}} \right)} \right] = \lim\limits_{n \to \infty } \left[ {\frac{1}{x}\ln {{\left( {1 + \frac{1}{n}} \right)}^n}} \right].\]
The factor \({\frac{1}{x}}\) does not depend on \(n\) and it can be taken out of the limit sign. Applying the composition limit law, we obtain:
\[y'\left( x \right) = \frac{1}{x}\lim\limits_{n \to \infty } \left[ {\ln {{\left( {1 + \frac{1}{n}} \right)}^n}} \right] = \frac{1}{x}\,\ln \left[ {\lim\limits_{n \to \infty } {{\left( {1 + \frac{1}{n}} \right)}^n}} \right].\]
As can be seen, the limit in the square brackets is equal to the number \(e\). Hence, the derivative of the natural logarithm is given by
\[y'\left( x \right) = \frac{1}{x}\ln e = \frac{1}{x}.\]
