Derivatives of Exponential Functions
Solved Problems
Example 7.
\[y = {2^{ - {x^2}}}\]
Solution.
We find the derivative using the chain rule.
\[y^\prime = \left( {{2^{ - {x^2}}}} \right)^\prime = {2^{ - {x^2}}}\ln 2 \cdot \left( { - {x^2}} \right)^\prime = {2^{ - {x^2}}}\ln 2 \cdot \left( { - 2x} \right) = - x{2^{1 - {x^2}}}\ln 2.\]
Example 8.
\[y = {4^x} \cdot {3^{2x}}\]
Solution.
Using the product and chain rules, we have:
\[y'\left( x \right) = \left( {{4^x} \cdot {3^{2x}}} \right)^\prime = {\left( {{4^x}} \right)^\prime } \cdot {3^{2x}} + {4^x} \cdot {\left( {{3^{2x}}} \right)^\prime } = {4^x}\ln 4 \cdot {3^{2x}} + {4^x} \cdot {3^{2x}}\ln 3 \cdot {\left( {2x} \right)^\prime } = {4^x}\ln 4 \cdot {3^{2x}} + {4^x} \cdot {3^{2x}}\ln 3 \cdot 2 = {4^x} \cdot {3^{2x}}\left( {\ln 4 + 2\ln 3} \right) = {4^x} \cdot {3^{2x}}\left( {\ln 4 + \ln {3^2}} \right) = {4^x} \cdot {3^{2x}}\left( {\ln 4 + \ln 9} \right) = {4^x} \cdot {3^{2x}}\ln 36.\]
There is also an easier way to calculate the derivative. First we simplify the given function by writing it as
\[y = {4^x} \cdot {3^{2x}} = {4^x} \cdot {\left( {{3^2}} \right)^x} = {4^x} \cdot {9^x} = {\left( {4 \cdot 9} \right)^x} = {36^x}.\]
Then the derivative is
\[y'\left( x \right) = {\left( {{4^x} \cdot {3^{2x}}} \right)^\prime } = {\left( {{{36}^x}} \right)^\prime } = {36^x}\ln 36.\]
Example 9.
\[y = {x^2}{2^x}.\]
Solution.
We use the product rule:
\[y^\prime = \left( {{x^2}{2^x}} \right)^\prime = \left( {{x^2}} \right)^\prime \cdot {2^x} + {x^2} \cdot \left( {{2^x}} \right)^\prime = 2x \cdot {2^x} + {x^2} \cdot {2^x}\ln 2 = x{2^x}\left( {2 + x\ln 2} \right).\]
Example 10.
\[y = \frac{{{x^2}}}{{{2^x}}}\]
Solution.
Using the quotient rule, we find:
\[\require{cancel} y'\left( x \right) = \left( {\frac{{{x^2}}}{{{2^x}}}} \right)^\prime = \frac{{{{\left( {{x^2}} \right)}^\prime } \cdot {2^x} - {x^2} \cdot {{\left( {{2^x}} \right)}^\prime }}}{{{{\left( {{2^x}} \right)}^2}}} = \frac{{2 x \cdot {2^x} - {x^2} \cdot {2^x}\ln 2}}{{{{\left( {{2^x}} \right)}^2}}} = \frac{{x\cancel{2^x}\left( {2 - x\ln 2} \right)}}{{{{\left( {{2^x}} \right)}^{\cancel{2}}}}} = \frac{{x\left( {2 - x\ln 2} \right)}}{{{2^x}}}.\;\;\]
Example 11.
\[y = \frac{{{x^n}}}{{{n^x}}}\;\left( {n \gt 0} \right)\]
Solution.
Here the numerator and denominator contain, respectively, a power and an exponential function. Using the quotient rule, we obtain:
\[y'\left( x \right) = \left( {\frac{{{x^n}}}{{{n^x}}}} \right)^\prime = \frac{{{{\left( {{x^n}} \right)}^\prime } \cdot {n^x} - {x^n} \cdot {{\left( {{n^x}} \right)}^\prime }}}{{{{\left( {{n^x}} \right)}^2}}} = \frac{{n {x^{n - 1}} \cdot {n^x} - {x^n} \cdot {n^x}\ln n}}{{{{\left( {{n^x}} \right)}^2}}} = \frac{{{x^{n - 1}}\cancel{n^x}\left( {n - x\ln n} \right)}}{{{{\left( {{n^x}} \right)}^{\cancel{2}}}}} = \frac{{{x^{n - 1}}\left( {n - x\ln n} \right)}}{{{n^x}}}.\]
Example 12.
\[y =\frac{{{e^x} - 1}}{{{e^x} + 1}}\]
Solution.
We use here the quotient rule:
\[y^\prime = \left( {\frac{{{e^x} - 1}}{{{e^x} + 1}}} \right)^\prime = \frac{{{e^x}\left( {{e^x} + 1} \right) - \left( {{e^x} - 1} \right){e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}} = \frac{{{\cancel{e^{2x}}} + {e^x} - \cancel{e^{2x}} + {e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}} = \frac{{2{e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}}.\]
Example 13.
\[y = {e^x}\left( {{x^2} - 2x + 2} \right)\]
Solution.
Using the product rule and the power rule, we have
\[y^\prime = \left[ {{e^x}\left( {{x^2} - 2x + 2} \right)} \right]^\prime = \left( {{e^x}} \right)^\prime \cdot \left( {{x^2} - 2x + 2} \right) + {e^x} \cdot \left( {{x^2} - 2x + 2} \right)^\prime = {e^x}\left( {{x^2} - 2x + 2} \right) + {e^x}\left( {2x - 2} \right) = {e^x}\left( {\color{darkgreen}{x^2} - \cancel{\color{blue}{2x}} + \cancel{\color{red}{2}} + \cancel{\color{blue}{2x}} - \cancel{\color{red}{2}}} \right) = {e^x}\color{darkgreen}{x^2}.\]
Example 14.
\[y = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}\]
Solution.
Combining the quotient rule and the chain rule, we get:
\[y^\prime = \left( {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right)^\prime = \frac{{\left( {{e^x} + {e^{ - x}}} \right)\left( {{e^x} + {e^{ - x}}} \right) - \left( {{e^x} - {e^{ - x}}} \right)\left( {{e^x} - {e^{ - x}}} \right)}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} = \frac{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2} - {{\left( {{e^x} - {e^{ - x}}} \right)}^2}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} = \frac{{\cancel{e^{2x}} + 2 + \cancel{e^{ - 2x}} - \cancel{e^{ - 2x}} + 2 - \cancel{e^{ - 2x}}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} = \frac{4}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}.\]
Example 15.
Show that the function \(y = {e^x}\left( {\sin 2x + \cos 2x} \right)\) is a solution of the differential equation
\[y^{\prime\prime} - 2y' + 5y = 0.\]
Solution.
Calculate the derivatives \(y'\) and \(y^{\prime\prime}:\)
\[y' = \left[ {{e^x}\left( {\sin 2x + \cos 2x} \right)} \right]^\prime = {\left( {{e^x}} \right)^\prime }\left( {\sin 2x + \cos 2x} \right) + {e^x}{\left( {\sin 2x + \cos 2x} \right)^\prime } = {e^x}\left( {\color{blue}{\sin 2x} + \color{red}{\cos 2x}} \right) + {e^x}\left( {\color{red}{2\cos 2x} - \color{blue}{2\sin 2x}} \right) = {e^x}\left( {\color{red}{3\cos 2x} - \color{blue}{\sin 2x}} \right);\]
\[y^{\prime\prime} = \left[ {{e^x}\left( {3\cos 2x - \sin 2x} \right)} \right]^\prime = {\left( {{e^x}} \right)^\prime }\left( {3\cos 2x - \sin 2x} \right) + {e^x}{\left( {3\cos 2x - \sin 2x} \right)^\prime } = {e^x}\left( {\color{red}{3\cos 2x} - \color{blue}{\sin 2x}} \right) + {e^x}\left( { - \color{blue}{6\sin 2x} - \color{red}{2\cos 2x}} \right) = {e^x}\left( {\color{red}{\cos 2x} - \color{blue}{7\sin 2x}} \right).\]
Substitute the derivatives and the function itself into the equation:
\[y^{\prime\prime} - 2y' + 5y = 0,\;\; \Rightarrow {e^x}\left( {\color{red}{\cos 2x} - \color{blue}{7\sin 2x}} \right) - 2{e^x}\left( {\color{red}{3\cos 2x} - \color{blue}{\sin 2x}} \right) + 5{e^x}\left( {\color{blue}{\sin 2x} + \color{red}{\cos 2x}} \right) = 0,\;\; \Rightarrow \color{red}{\cos 2x} - \color{blue}{7\sin 2x} - \color{red}{6\cos 2x} + \color{blue}{2\sin 2x} + \color{blue}{5\sin 2x} + \color{red}{5\cos 2x} = \color{black}{0},\;\; \Rightarrow 0 \equiv 0.\]
Hence, this function is a solution of the differential equation.
Example 16.
Show that the function \(y = {e^{ - 2x}}\left( {\sin 3x + \cos 3x} \right)\) is a solution of the differential equation
\[y^{\prime\prime} + 4y' + 13y = 0.\]
Solution.
First we determine the first and second order derivatives of the function:
\[y' = \left[ {{e^{ - 2x}}\left( {\sin 3x + \cos 3x} \right)} \right]^\prime = {\left( {{e^{ - 2x}}} \right)^\prime }\left( {\sin 3x + \cos 3x} \right) + {e^{ - 2x}}{\left( {\sin 3x + \cos 3x} \right)^\prime } = - 2{e^{ - 2x}}\left( {\color{blue}{\sin 3x} + \color{red}{\cos 3x}} \right) + {e^{ - 2x}}\left( {\color{red}{3\cos 3x} - \color{blue}{3\sin 3x}} \right) = {e^{ - 2x}}\left( {\color{red}{\cos 3x} - \color{blue}{5\sin 3x}} \right);\]
\[y^{\prime\prime} = \left[ {{e^{ - 2x}}\left( {\cos 3x - 5\sin 3x} \right)} \right]^\prime = {\left( {{e^{ - 2x}}} \right)^\prime }\left( {\cos 3x - 5\sin 3x} \right) + {e^{ - 2x}}{\left( {\cos 3x - 5\sin 3x} \right)^\prime } = - 2{e^{ - 2x}}\left( {\color{red}{\cos 3x} - \color{blue}{5\sin 3x}} \right) + {e^{ - 2x}}\left( { - \color{blue}{3\sin 3x} - \color{red}{15\cos 3x}} \right) = {e^{ - 2x}}\left( {\color{blue}{7\sin 3x} - \color{red}{17\cos 3x}} \right).\]
Substituting the derivatives and the function into the equation yields:
\[y^{\prime\prime} + 4y' + 13y = 0,\;\; \Rightarrow {e^{ - 2x}}\left( {\color{blue}{7\sin 3x} - \color{red}{17\cos 3x}} \right) + 4{e^{ - 2x}}\left( {\color{red}{\cos 3x} - \color{blue}{5\sin 3x}} \right) + 13{e^{ - 2x}}\left( {\color{blue}{\sin 3x} + \color{red}{\cos 3x}} \right) = 0,\;\; \Rightarrow \color{blue}{7\sin 3x} - \color{red}{17\cos 3x} + \color{red}{4\cos 3x} - \color{blue}{20\sin 3x} + \color{blue}{13\sin 3x} + \color{red}{13\cos 3x} = \color{black}{0},\;\; \Rightarrow 0 \equiv 0.\]
So, the function is a solution of the differential equation.
