Differentiation of Exponential and Logarithmic Functions: Definitions, Differentiation Formulas, Solved Examples

Differentiation of Exponential and Logarithmic Functions: In nature, there are various processes where the growth or decay of a quantity has unusual qualities such as rate of decay of radioactive materials, rate of growth of some bacteria, etc. These processes share a common feature known as exponential growth or decay. The functions that express them are known as exponential functions. On the other hand, the logarithmic scales condense large volumes into small areas. To indicate the rate of change in any function, we employ the concept of derivatives. This considers even a very small change in the dependent variable with respect to small changes in independent variables.

What are Logarithms?

Consider \({a^x} = N\). Here, we are expressing \(N\) in terms of \(a\) and \(x\). Another way to write the same relation is \(a = {N^{\frac{1}{x}}}\). Here, we are expressing \(a\) in terms of \(N\) and \(x\). If we must express \(x\) in terms of the other two parameters \(a\) and \(N\) then we have to use logarithms. This can be written as a logarithmic function as,
\({\log _a}N\)
It can be defined as, “The logarithm \(x\) of a positive number \(N\) to a positive base \(a\) is the index to which the base is raised to equal the given number.”
\({a^x} = N\)
\(\therefore x = {\log _a}N\)
Note:
1. In the equation, \(N\) is a positive number, and \(a\) is any positive number other than \(1\).
2. Unity cannot be the base of the logarithm, as \({\log _1}N\) is not defined. Also, \({\log _1}1\) is not unique. It has infinitely many solutions, which is necessary for functional notation.

Logarithmic Differentiation

Let \(y = {\log _a}x\,\,\,\,\,…….\left( 1 \right)\)
Here, \(a > 1\)
Let \(\delta x\) be a small increment in \(x\) and \(\delta y\) the corresponding increment in \(y\)
\(\therefore y + \delta y = {\log _a}\left( {x + \delta x} \right)\,\,\,\,……\left( 2 \right)\)
Subtracting \(\left( 1 \right)\) from \(\left( 2 \right)\) we have
\(\delta y = {\log _a}\left( {x + \delta x} \right) – {\log _a}x\)
\(\delta y = {\log _a}\left( {\frac{{x + \delta x}}{x}} \right) = {\log _a}\left( {1 + \frac{{\delta x}}{x}} \right)\,\,\,\,…..\left( 3 \right)\)
Dividing both sides of equation \(\left( 3 \right)\) by \(\delta x\), we have
\(\frac{{\delta y}}{{\delta x}} = \frac{1}{{\delta x}}{\log _a}\left( {1 + \frac{{\delta x}}{x}} \right)\)
\(\frac{{\delta y}}{{\delta x}} = \frac{1}{x} \cdot \frac{x}{{\delta x}}{\log _a}\left( {1 + \frac{{\delta x}}{x}} \right)\)
\(\frac{{\delta y}}{{\delta x}} = \frac{1}{x}{\log _a}{\left( {1 + \frac{{\delta x}}{x}} \right)^{\frac{x}{{\delta x}}}}\)
Proceeding to limits as \(\delta x \to 0\), we have
\(\mathop {\lim }\limits_{\delta x \to 0} \frac{{\delta y}}{{\delta x}} = \mathop {\lim }\limits_{\delta x \to 0} \left[ {\frac{1}{x}{{\log }_a}{{\left( {1 + \frac{{\delta x}}{x}} \right)}^{\frac{x}{{\delta x}}}}} \right]\)
\( \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}{\log _a}e\)
\(\left[ {\because \mathop {\lim }\limits_{\delta x \to 0} {{\left( {1 + \frac{{\delta x}}{x}} \right)}^{\frac{x}{{\delta x}}}} = e} \right]\)
Hence, \(\frac{d}{{dx}}{\log _a}x = \frac{1}{x}{\log _a}e\)
Note:

• \(\frac{d}{{dx}}{\log _e}x = \frac{1}{x}\)
• \(\frac{d}{{dx}}{a^x} = {a^x}{\log _e}a,x > 0\)

A set of logarithmic properties is given below, which helps simplify the functions to perform the differentiation process.

Applications of Logarithmic Differentiation

The logarithmic differentiation has two main applications. It helps to reduce the calculation for differentiation of functions.

1. Product of Functions

The application of logarithms changes the product of two or more functions into the sum of functions, allowing for easy differentiation of the function. Let the function \(f\left( x \right)\) be the product of two sub-functions \(g\left( x \right)\) and \(h\left( x \right)\), respectively, and we can apply logarithmic followed by differentiation of the functions.
\(f\left( x \right) = g\left( x \right) \cdot h\left( x \right)\)
Let’s use logarithms to express the product of functions on both sides of the equation above.
\(\log f\left( x \right) = \log \left( {g\left( x \right) \cdot h\left( x \right)} \right)\)
\( \Rightarrow \log f\left( x \right) = \log g\left( x \right) + \log h\left( x \right)\)
Now on differentiating both sides with respect to \(x\), we get
\(\frac{{d\left( {\log f\left( x \right)} \right)}}{{dx}} = \frac{{d\left( {\log g\left( x \right) + \log h\left( x \right)} \right)}}{{dx}}\)
\(\frac{1}{{f\left( x \right)}} \cdot {f^\prime }\left( x \right) = \frac{1}{{g\left( x \right)}} \cdot {g^\prime }\left( x \right) + \frac{1}{{h\left( x \right)}} \cdot {h^\prime }\left( x \right)\)
Where \({f^\prime }\left( x \right),{g^\prime }\left( x \right)\) and \({h^\prime }\left( x \right)\) are derivatives of the functions \(f\left( x \right),g\left( x \right)\) and \(h\left( x \right)\) respectively.
\(\frac{{{f^\prime }\left( x \right)}}{{f\left( x \right)}} = \frac{{h\left( x \right) \cdot {g^\prime }\left( x \right) + g\left( x \right) \cdot {h^\prime }\left( x \right)}}{{g\left( x \right)h\left( x \right)}}\)
\({f^\prime }\left( x \right) = f\left( x \right)\left[ {\frac{{h\left( x \right) \cdot {g^\prime }\left( x \right) + g\left( x \right) \cdot {h^\prime }\left( x \right)}}{{g\left( x \right)h\left( x \right)}}} \right]\)
\(\therefore {f^\prime }\left( x \right) = h\left( x \right){g^\prime }\left( x \right) + g\left( x \right){h^\prime }\left( x \right)\)

2. Division of Functions

The procedure of logarithmic differentiation is used to differentiate the division of one function by another function, commonly known as the quotient of functions. The division of one function by another is transformed into a difference in the logarithms of the two functions. Let us consider a function \(f\left( x \right)\), which is equal to the quotient of the two other functions \(g\left( x \right)\), and \(h\left( x \right)\).
\( \Rightarrow f\left( x \right) = \frac{{g\left( x \right)}}{{h\left( x \right)}}\)
On taking \(\log \) on both sides, we get
\(\log f\left( x \right) = \log \left( {\frac{{g\left( x \right)}}{{h\left( x \right)}}} \right)\)
\( \Rightarrow \log f\left( x \right) = \log g\left( x \right) – \log h\left( x \right)\)
Now on differentiating both sides with respect to \(x\), we get
\(\frac{{d\left( {\log f\left( x \right)} \right)}}{{dx}} = \frac{{d\left( {\log g\left( x \right) – \log h\left( x \right)} \right)}}{{dx}}\)
\(\frac{1}{{f\left( x \right)}} \cdot {f^\prime }\left( x \right) = \frac{1}{{g\left( x \right)}} \cdot {g^\prime }\left( x \right) – \frac{1}{{h\left( x \right)}} \cdot {h^\prime }\left( x \right)\)
Where \(f^\prime (x), g^\prime (x)\) and \(h^\prime (x)\) are derivatives of the functions \(f(x),\,g(x)\) and \(h(x)\), respectively.
\(\frac{{{f^\prime }\left( x \right)}}{{f\left( x \right)}} = \frac{{h\left( x \right){g^\prime }\left( x \right) – g\left( x \right){h^\prime }\left( x \right)}}{{g\left( x \right)h\left( x \right)}}\)
\(f’\left( x \right) = f\left( x \right)\left[ {\frac{{h\left( x \right)g’\left( x \right) – g\left( x \right)h’\left( x \right)}}{{g\left( x \right)h\left( x \right)}}} \right]\)
\(\therefore {f^\prime }\left( x \right) = h\left( x \right){g^\prime }\left( x \right) – g\left( x \right){h^\prime }\left( x \right)\)

Steps to Solve Logarithmic Differentiation

Here are the steps below to find the differentiation of functions using logarithmic differentiation
Step 1: Assume the given function as \(y = f\left( x \right)\)
Step 2: Take log on both sides of the given function.
Step 3: Use the log properties to remove the exponent.
Step 4: Differentiate the obtained equation.
Step 5: Simplify the equation for \(\frac{{dy}}{{dx}}\) .
Step 6: Substitute the value of \(y = f\left( x \right)\) in obtained equation \(\frac{{dy}}{{dx}}\).

What are Exponential Functions?

Consider two functions, one of which is an exponent of another. Using the application of logarithms, i.e., power rule changes it into a product of one function and the exponent of the other. Let us take a function \(f\left( x \right)\), which is equal to the exponent of \(g\left( x \right)\) to \(h\left( x \right)\).
\(f\left( x \right) = g{\left( x \right)^{h\left( x \right)}}\)
Apply logarithms on both sides,
\(\log f\left( x \right) = \log \left( {g{{\left( x \right)}^{h\left( x \right)}}} \right)\)
\(\log f\left( x \right) = h\left( x \right) \cdot \log \left( {g\left( x \right)} \right)\)
Now on differentiating both the sides with respect to \(x\), we get
\(\frac{{d\left( {\log f\left( x \right)} \right)}}{{dx}} = \frac{{d(h\left( x \right) \cdot \log \left( {g\left( x \right)} \right)}}{{dx}}\)
\(\frac{1}{{f\left( x \right)}} \cdot {f^\prime }\left( x \right) = {h^\prime }\left( x \right)\log g\left( x \right) + h\left( x \right)\frac{1}{{g\left( x \right)}}{g^\prime }\left( x \right)\)
\(\frac{{{f^\prime }\left( x \right)}}{{f\left( x \right)}} = \frac{{\left[ {g\left( x \right){h^\prime }\left( x \right)\log g\left( x \right) + h\left( x \right){g^\prime }\left( x \right)} \right]}}{{g\left( x \right)}}\)
\({f^\prime }(x) = f\left( x \right)\left[ {\frac{{\left[ {g\left( x \right){h^\prime }\left( x \right)\log g\left( x \right) + h\left( x \right){g^\prime }\left( x \right)} \right]}}{{g\left( x \right)}}} \right]\)
On putting the value of \({f^\prime }(x) = f’\left( x \right)\), we get
\({f^\prime }\left( x \right) = g{\left( x \right)^{h\left( x \right)}}\left[ {\frac{{\left[ {g\left( x \right){h^\prime }\left( x \right)\log g\left( x \right) + h\left( x \right){g^\prime }(x)} \right]}}{{g\left( x \right)}}} \right]\)
\( \Rightarrow {f^\prime }\left( x \right) = g{\left( x \right)^{h\left( x \right) – 1}}\left[ {g\left( x \right){h^\prime }\left( x \right)\log g\left( x \right) + h\left( x \right){g^\prime }\left( x \right)} \right]\)
Now let us consider \(f\left( x \right) = {a^x}\). The power rule, which we examined a few sections ago, won’t function because the exponent had to be a constant integer and the base had to be a variable. That is the polar opposite of what we have with this function. As a result, we’ll have to start with the derivative’s definition.
\({f^\prime }\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) – f\left( x \right)}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{{a^{x + h}} – {a^x}}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{{a^x}{a^h} – {a^x}}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{{a^x}\left( {{a^h} – 1} \right)}}{h}\)
\({a^x}\) is not affected by the limit because it doesn’t have any \(h\) in it. It is a constant as far as the limit is concerned. Now we can take the factor out of the limit.
\(\therefore {f^\prime }\left( x \right) = {a^x}\mathop {\lim }\limits_{h \to 0} \frac{{{a^h} – 1}}{h}\)
Now, notice that the limit we have got above is precisely the definition of the derivative of \(f\left( x \right) = {a^x}\) at \(x = 0\), i.e., \({f^\prime }\left( 0 \right)\). Therefore, the derivative becomes,
\({f^\prime }\left( x \right) = {f^\prime }\left( 0 \right){a^x}\)

Example of Exponential Functions:

Below are few examples,
1. \(f\left( x \right) = {x^{\cos x}}\)
2. \(g\left( x \right) = \tan {x^x}\)
3. \(h\left( x \right) = {e^{x\log x}}\)

PRACTICE EXAM QUESTIONS AT EMBIBE

Solved Examples – Differentiation of Exponential and Logarithmic Functions

Below are a few solved examples that can help in getting a better idea.

Q.1. Differentiate \(y = \frac{{{x^{\frac{1}{2}}}{{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{5}}}}}\) with respect to \(x\).
Ans: Given: \(y = \frac{{{x^{\frac{1}{2}}}{{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{5}}}}}\)
Taking logarithm both sides, we obtain
\(\log y = \frac{1}{2}\log x + \frac{2}{3}\log \left( {1 – 2x} \right) – \frac{3}{4}\log \left( {2 – 3x} \right) – \frac{4}{5}\log \left( {3 – 4x} \right)\)
Differentiating with respect to \(x\) we get
\(\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \frac{1}{2} \cdot \frac{1}{x} + \frac{2}{3} \cdot \left( {\frac{{ – 2}}{{1 – 2x}}} \right) – \frac{3}{4}\left( {\frac{{ – 3}}{{2 – 3x}}} \right) – \frac{4}{5}\left( {\frac{{ – 4}}{{3 – 4x}}} \right)\)
\(\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{5\left( {3 – 4x} \right)}}\)
\( \Rightarrow \frac{{dy}}{{dx}} = y\left[ {\frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{5\left( {3 – 4x} \right)}}} \right]\)
Hence, \(\frac{{dy}}{{dx}} = \left( {\frac{{{x^{\frac{1}{2}}}{{\left( {1 – 2x} \right)}^{\frac{2}{3}}}}}{{{{\left( {2 – 3x} \right)}^{\frac{3}{4}}}{{\left( {3 – 4x} \right)}^{\frac{4}{5}}}}}} \right)\left[ {\frac{1}{{2x}} – \frac{4}{{3\left( {1 – 2x} \right)}} + \frac{9}{{4\left( {2 – 3x} \right)}} + \frac{{16}}{{5\left( {3 – 4x} \right)}}} \right]\)

Q.2. Find the derivative of \(y = {x^{\sin x}}\).
Ans: Given: \(y = {x^{\sin x}}\)
Taking logarithm both sides, we obtain
\(\log y = \sin x\log x\)
Differentiating with respect to \(x\), we get
\(\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \cos x\log x + \sin x \cdot \frac{1}{x}\)
\( \Rightarrow \frac{{dy}}{{dx}} = y\left[ {\cos x\log x + \frac{{\sin x}}{x}} \right]\)
Hence, \(\frac{{dy}}{{dx}} = {x^{\sin x}}\left[ {\cos x\log x + \frac{{\sin x}}{x}} \right]\)

Q.3. Find \(\frac{{dy}}{{dx}}\) of \({x^y} = {y^x}\)
Ans: Given \({x^y} = {y^x}\), taking logarithm both the sides, we get
\(\log {x^y} = \log {y^x}\)
\( \Rightarrow y\log x = x\log y\)
Now differentiate with respect to \(x\), we get
\(\frac{{dy}}{{dx}}\log x + \frac{1}{x}y = \frac{x}{y}\left( {\frac{{dy}}{{dx}}} \right) + \log y\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {\log x – \frac{x}{y}} \right) = \log y – \frac{y}{x}\)
\( \Rightarrow \frac{{dy}}{{dx}} = \left[ {\frac{{\left( {\log y – \frac{y}{x}} \right)}}{{\left( {\log x – \frac{x}{y}} \right)}}} \right]\)
Hence, \(\frac{{dy}}{{dx}} = \frac{y}{x} \cdot \left( {\frac{{x\log y – y}}{{y\log x – x}}} \right)\)

Q.4. Find \(\frac{{dy}}{{dx}}\) for \(y = {e^{{x^{{e^x}}}}} + {x^{{e^{{e^x}}}}}\)
Ans: Let \(y = u + v\)
\(u = {e^{{x^{{e^x}}}}}\) taking logarithm both the sides, we get
\(\log u = {x^{{e^x}}}\log e = \log u = {x^{{e^x}}}\)
Again taking logarithm both the sides, we get
\(\log \log u = {e^x}\log x\)
Differentiating with respect to \(x\), we get
\(\frac{1}{{\log u}} \times \frac{1}{u}\left( {\frac{{du}}{{dx}}} \right) = {e^x}\frac{1}{x} + {e^x}\log x\)
\( \Rightarrow \frac{{du}}{{dx}} = u\log u\left( {\frac{{{e^x}}}{x} + {e^x}\log x} \right)\)
\( \Rightarrow \frac{{du}}{{dx}} = {e^{{x^{{e^x}}}}}{x^{{e^x}}}{e^x}\left( {\log x + \frac{1}{x}} \right)\)
Now \(v = {x^{{e^{{e^x}}}}}\) taking logarithm both the sides, we get
\(\log v = {e^{{e^x}}}\log x\)
Again taking logarithm both the sides, we get
\(\log \log v = \log {e^{{e^x}}} + \log \log x\)
\( \Rightarrow \log \log v = {e^x} + \log \log x\)
Differentiating with respect to \(x\), we get
\(\frac{1}{{\log v}} \times \frac{1}{v}\frac{{dv}}{{dx}} = {e^x} + \frac{1}{{\log x}}\frac{1}{x}\)
\( \Rightarrow \frac{{dv}}{{dx}} = v\log v\left( {{e^x} + \frac{1}{{x\log x}}} \right)\)
\( \Rightarrow \frac{{dv}}{{dx}} = {x^{{e^{{e^x}}}}}{e^{{e^x}}}\log x\left( {{e^x} + \frac{1}{{x\log x}}} \right)\)
Now \(\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}\)
Hence, \(\frac{{dy}}{{dx}} = {e^{{x^{{e^x}}}}}{x^{{e^x}}}{e^x}\left( {\log x + \frac{1}{x}} \right) + {x^{{e^{{e^x}}}}}{e^{{e^x}}}\log x\left( {{e^x} + \frac{1}{{x\log x}}} \right)\)

Q.5. Find the derivative of \({x^2}\sin x{e^x}\)
Ans: Let \(y = {x^2}\sin x{e^x}\)
Now taking logarithm both sides, we get
\(\log y = \log \left( {{x^2}\sin x{e^x}} \right)\)
\( \Rightarrow \log y = \log {x^2} + \log \sin x + \log {e^x}\)
\( \Rightarrow \log y = 2\log x + \log \sin x + x\)
Differentiating with respect to \(x\), we get
\(\frac{1}{y}\frac{{dy}}{{dx}} = \frac{2}{x} + \frac{1}{{\sin x}}\left( {\cos x} \right) + 1\)
\( \Rightarrow \frac{{dy}}{{dx}} = y\left( {\frac{2}{x} + \cot x + 1} \right)\)
Hence, \(\frac{{dy}}{{dx}} = {x^2}\sin x{e^x}\left( {\frac{2}{x} + \cot x + 1} \right)\)

Summary

The logarithm can be defined as, “The logarithm \(x\) of a positive number \(N\) to a positive base \(a\) is the index to which the base is raised in order to equal the given number.”\({a^x} = N \Rightarrow x = \log aN\) . Differentiation of logarithm is given by \(\frac{d}{{dx}}\log_a x = \frac{1}{x}{\log _a}e\) . Logarithmic differentiation has two main applications product of functions and quotient of functions. It helps to reduce the calculation for differentiation of functions. For two functions, one is an exponent of another function. The application of logarithms to this function transforms it into a product of one function and the exponent of the other function. The differentiation of the exponential function is given by \({f^\prime }\left( x \right) = g{\left( x \right)^{h\left( x \right) – 1}}\left[ {g\left( x \right){h^\prime }\left( x \right)\log g\left( x \right) + h\left( x \right){g^\prime }\left( x \right)} \right]\).

Frequently Asked Questions (FAQs)

Students might be having many questions regarding the Differentiation of Exponential and Logarithmic Functions. Here are a few commonly asked questions and answers.

Q.1. What is the logarithmic differentiation formula?
Ans: For a logarithmic function \(y = \log_a x\) the derivative is given by \(\frac{d}{{dx}}\log_a x = \frac{1}{x}{\log _a}e\).

Q.2. Is logarithmic differentiation the same as implicit differentiation?
Ans: No, both are not the same. They have different properties and types of functions for which we use this differentiation. We use logarithmic differentiation when we have expressions of the form \(y = f{\left( x \right)^{g\left( x \right)}}\) i.e., a variable to the power of a variable. We can not apply the power rule and the exponential here, and we use implicit differentiation when the function \(y\) is only implicitly defined. When it is impossible to solve and get \(y = …,\), we use implicit differentiation.

Q.3. What are logarithmic and exponential functions?
Ans: Logarithmic Function: The logarithm can be defined as, “The logarithm \(x\) of a positive number \(N\) to a positive base \((a)\) is the index to which the base is raised in order to equal the given number.” \({a^x} = N \Rightarrow x = {\log _a}N\).
Exponential Function: A relationship of the type \(y = {a^x}\) Where the independent variable \(x\) is the exponent of a positive number \(a\) and ranges over the entire real number line. The exponential function \(y = {e^x}\), sometimes written \(y = \exp \left( x \right)\), in which \(e\left( {2.7182818 \ldots } \right)\) is the basis of the natural system of logarithms \(\ln \), which is probably the most significant.

Q.4. How would you differentiate between exponential and logarithmic functions?
Ans: \(f\left( x \right) = {e^x}\) denotes the exponential function, whereas \(g\left( x \right) = {\log _e}x\) denotes the logarithmic function, and both are inverse of each other. The exponential function’s domain is a set of real numbers, whereas the logarithmic function’s domain is a set of positive real numbers.

Q.5. What is the differentiation of exponential function?
Ans: For two functions such that one is an exponent of another function, the application of logarithms to this function transforms it into a product of one function and the exponent of the other function. Let us take a function \(f\left( x \right)\), which is equal to the exponent of \(g\left( x \right)\) to \(h\left( x \right)\), i.e., \(f\left( x \right) = g{\left( x \right)^{h\left( x \right)}}\). The differentiation of the exponential function is given by \({f^\prime }\left( x \right) = g{\left( x \right)^{h\left( x \right) – 1}}\left[ {g\left( x \right){h^\prime }\left( x \right)\log g\left( x \right) + h\left( x \right){g^\prime }\left( x \right)} \right]\).

ATTEMPT MOCK TESTS ON EMBIBE

We hope this information about the Differentiation of Exponential and Logarithmic Functions has been helpful. If you have any doubts, comment in the section below, and we will get back to you.