Differentiation Of Exponential And Logarithmic Functions

In this article, we will learn about Differentiation of Exponential and Logarithmic Functions. The article is written as per the requirements of students of Secondary 4 A Maths grade. In this topic, we will learn about:

• Derivative of \(e^x\)
• Derivative of \(e^{f(x)}\)
• Derivative of \(log \;x\)
• Derivative of \(log \;f(x)\)

Differentiating \(e^x\)

When you differentiate the most basic exponential function, \(e^x\), you get \(e^x\).

\(\begin{align*} \frac {d} {dx}e^x = e^x \end{align*}\)

Here, ‘\(e\)’ is 2.718. This number is calculated in such a way, so that \(\begin{align*} \frac {d} {dx}\;(e^x) = e^x \end{align*}\) is obtained.

Example 1:

Differentiate each of the following with respect to \(x\).

1. \(\begin{align*} 2xe^x \end{align*}\)
2. \(\begin{align*} e^x \; cos^2(2x) \end{align*}\), where \(x\) is in radians.
3. \(\begin{align*} \frac {e^x-1 } {e^x+1 } \end{align*}\)

Solution: 

1. When we differentiate, we keep the first term and differentiate the second term. We add to the second term and differentiate it too.

We will solve it by the product rule.

\(\begin{align*} \frac {d} {dx} (2xe^x) &= 2x\frac {d} {dx} (e^x) + e^x\frac {d} {dx} (2x) \\ \\ &= 2xe^x + 2e^x \\ \\ &= 2e^x(x+1) \end{align*}\)

2. We will solve it by the product rule. 

\(\begin{align*} \frac {d} {dx} [cos^2(2x)] &= e^x\frac {d} {dx} [cos\;(2x)]^2 + cos^22x \frac{d}{dx}(e^x) \\ \\ &= e^x[2(cos\;2x)^1(-sin\;2x)(2)]+e^xcos^22x \\ \\ &= -4e^x\;sin\;2xcos\;2x\; +\; e^xcos^22x \end{align*}\)

3. We will use the quotient rule to solve the problem. 

\(\begin{align*} \frac{d}{dx}\big[\;\frac {e^x\;-\;1 } {e^x\;+\;1 }\;\big] &= \frac {(e^x+1)\;\frac {d}{dx}(e^x-1)-(e^x-1)\;\frac{d}{dx}(e^x+1)} {(e^x+1)^2} \\ \\ &=\frac {(e^x+1)(e^x)-(e^x-1)(e^x)} {(e^x+1)^2} \\ \\ &= \frac {e^x(e^x+1-e^x+1)} {(e^x+1)^2} \\ \\ &= \frac {2e^x} {(e^x+1)^2} \end{align*} \)

Chain Rule On Exponential Functions

In a chain rule, if an exponent has an outer function and an inner function, the outer function must be differentiated and multiplied by the derivative of the inner function.

\(\begin{align*} &1.\; \frac {d}{dx} [e^{(f(x)}] = e^{f(x)} \;[f'(x)] = f'(x)e^{f(x)} \\ \\ &2.\; \frac {d}{dx} (e^{ax+b}) = ae^{ax+b} \end{align*}\)

Example 2:

Differentiate each of the following with respect to \(x\), where \(x\) is in radians. 

\(\begin{align*} &1.{ \;e^{-x}} \\ \\ &2. {\;e^{\;4x-1}\; e^{\sqrt {x}} \over e^x} \\ \\ &3. {\;e^{3x}\;sin\,3x} \\ \\ &4. {\;e^{\;3x+2} \over x } \end{align*} \)

Solution:

1.  \(\begin{align*} \frac {d} {dx}\;e^{-x} = -e^{-x} \end{align*}\)

2. We will use the Law of Indices. 

    \(\begin{align*} \frac {d} {dx}\;\bigg (\frac { e^{\;4x\,-\,1} \; e^{\sqrt{x}} } {e^x} \bigg) &= \frac {d} {dx} \bigg ( e^{(4x\,-\,1)\;+\;\sqrt {x} \,- \,x} \bigg) \\ \\ &= \frac {d} {dx} \bigg ( e^{3x\,-\,1\,+\,x^{\frac {1}{2}}} \bigg) \end{align*}\)

    We will then use the differentiating exponent through chain rule. 

\(\begin{align*} &= \textstyle e^{ 3x-1+x^{ \frac {1}{2}} }\frac{d}{dy} (3x-1+x^{\frac {1}{2}}) \\ \\ &= \textstyle e^{ 3x+1+\sqrt {x} } ( 3+ \frac {1}{2}x^{-\frac {1}{2}} ) \\ \\ &= \textstyle e^{ 3x+1+\sqrt {x} } ( 3+ \frac {1}{2\sqrt {x} } ) \\ \\ &=\textstyle ( 3+ \frac {1}{2\sqrt {x} } )e^{ 3x-1+\sqrt {x} } \end{align*}\)

3.  We use the product rule to solve the problem. 

    \(\begin{align*} \frac {d}{dx} [e^{3x} sin(3x)] &=3e^{3x}cos3x+3sin3x \;e^{3x} \\ \\ &= 3e^{3x} (cos3x+sin3x) \end{align*}\)

4.  We will use the quotient rule to solve the problem. 

     \(\begin{align*} \frac {d}{dx} \bigg[\frac {e^{3x+2}}{x}\bigg] &=\frac {x(3e^{3x+2} )- e^{3x+2}}{x^2} \\ \\ &= \frac {e^{3x+2}(3x-1)}{x^2} \end{align*}\)

Differentiating \(\ln x\)

Differentiating \(\ln x\) is simple and gives:

\(\begin{align*} \frac {d} {dx}\;(\ln\;x) = \frac {1}{x} \end{align*}\)

Example 3:

Differentiate each of the following with respect to \(x\). 

1. \(\begin{align*} x \; \ln\; x \end{align*}\)
2. \(\begin{align*} \frac {\ln\;x}{2x} \end{align*}\)

Solution:

1. We use the product rule to solve the problem. 

\(\begin{align*}​​ \frac {d}{dx} ( x\;\ln\;x) &=​​\textstyle x(\frac {1}{x}) \;+\; \ln\;x(1) \\ \\ &= 1\;+\; \ln\;x \end{align*}\)

2. We use the quotient rule to solve the problem. 

\(\begin{align*}\frac {d}{dx} \bigg( \frac {\ln x}{2x}\bigg) &= \frac {2x(\frac {1}{x})-2\ln x }{4x^2} \\ \\ &= \frac {2(1-\ln x)}{4x^2} \\ \\ &=\frac {1-\ln x}{2x^2} \end{align*}\)

Chain Rule On Logarithmic Functions

After differentiating \(\ln x\), let us now look at how to differentiate  \(\ln f(x)\).

\(\begin{align*} \frac {d}{dx}\;\ln f(x) &=\frac {1}{f(x)} f'(x) \\ \\ &=\frac {f(x)}{f'(x)} \end{align*}\)

Example 4:

Differentiate \(\begin{align*} \ln x² \end{align*}\).

Solution:

Note:

\(\ln x^2= \ln (x^2) \) and not \( (\ln x)^2 \)

\(\begin{align*} \frac {d}{dx}\;(\ln x^2) &= \textstyle\frac {2x}{x^2}\\ \\ &=\textstyle \frac {2}{x} \end{align*} \)

Conclusion

In the article, we learned about Differentiation of Exponential and Logarithmic Functions as per the syllabus of the Secondary 4 A Maths class. We discussed about differentiating the exponential function \(e^x\). We also studied about chain rules on exponential functions and differentiating \(\begin{align*} \ln x \end{align*}\).