Exponential and Logarithmic Differentiation and Integration have a lot of practical applications and are handled a little differently than we are used to. For a review of these functions, visit the Exponential Functions section and the Logarithmic Functions section.
Note that we will address Exponential and Logarithmic Integration here in the Integration section.
Introduction to Exponential and Logarithmic Differentiation and Integration
Before getting started, here is a table of the most common Exponential and Logarithmic formulas for Differentiation and Integration:
| Exponential and Logarithmic Derivatives | Exponential and Logarithmic Integrals | Examples |
| $ \displaystyle \frac{d}{{dx}}\left( {\ln u} \right)=\frac{{{u}’}}{u}$ | $ \displaystyle \int{{\frac{{{u}’}}{u}}}\,dx=\ln \left| u \right|\,+C$ (We need absolute value since $ \ln u$ is only defined for $ u>0$ .) | $ \displaystyle \color{#800000}{{\frac{{d\left[ {\ln \left( {{{x}^{5}}-3} \right)} \right]}}{{dx}}}}=\frac{{{u}’}}{u}=\frac{{5{{x}^{4}}}}{{{{x}^{5}}-3}}\,\,\,\,\,\,\,\text{(}u={{x}^{5}}-3)$
$ \displaystyle \begin{align}u&={{x}^{5}}-3\\du&=5{{x}^{4}}\,dx\\dx&=\frac{{du}}{{5{{x}^{4}}}}\end{align}$ $ \require{cancel} \begin{align} \color{#800000}{{\int{{\frac{{5{{x}^{4}}}}{{{{x}^{5}}-3}}}}\,dx}}&=\int{{\frac{{5{{x}^{4}}}}{u}}}\,dx=\int{{\frac{{\cancel{{5{{x}^{4}}}}}}{u}}}\,\cdot \frac{{du}}{{\cancel{{5{{x}^{4}}}}}}\\&=\int{{\frac{{du}}{u}}}\,=\ln \left| u \right|+C=\ln \left| {{{x}^{5}}-3} \right|+C\end{align}$
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| $ \displaystyle \frac{d}{{dx}}\left( {{{{\log }}_{a}}u} \right)=\frac{{{u}’}}{{u\left( {\ln \,a} \right)}}$ | $ \displaystyle \int{{\frac{{{u}’}}{{u\left( {\ln a} \right)}}}}\,dx={{\log }_{a}}\left| u \right|\,+C$ (Not a typical integration problem) | $ \displaystyle \color{#800000}{{\frac{d}{{dx}}\left[ {\log \left( {4x-1} \right)} \right]}}=\frac{d}{{dx}}\left[ {{{{\log }}_{{10}}}\left( {4x-1} \right)} \right]=\frac{4}{{\left( {4x-1} \right)\ln 10}}$ |
| $ \displaystyle \frac{d}{{dx}}\left( {{{e}^{u}}} \right)={{e}^{u}}{u}’$ | $ \displaystyle \int{{{{e}^{u}}}}={{e}^{u}}+C$ | $ \displaystyle \color{#800000}{{\frac{{d\left( {{{e}^{{3x}}}} \right)}}{{dx}}}}={{e}^{{3x}}}\cdot 3=3{{e}^{{3x}}}$ $ \begin{align}u&=3x\\du&=3\,dx\\dx&=\frac{{du}}{3}\end{align}$ $ \displaystyle \color{#800000}{{\int{{3{{e}^{{3x}}}}}dx}}=3\int{{{{e}^{u}}dx}}=\cancel{3}\int{{{{e}^{u}}\cdot \frac{{du}}{{\cancel{3}}}}}={{e}^{u}}+C={{e}^{{3x}}}+C$ |
| $ \displaystyle \frac{d}{{dx}}\left( {{{a}^{u}}} \right)=\left( {\ln \,a} \right){{a}^{u}}{u}’$ | $ \displaystyle \int{{{{a}^{u}}}}du=\left( {\frac{1}{{\ln a}}} \right){{a}^{u}}+C$ | $ \displaystyle \color{#800000}{{\frac{d}{{dx}}\left( {{{4}^{{3{{x}^{3}}}}}} \right)}}=\ln 4\left( {{{4}^{{3{{x}^{3}}}}}} \right)\left( {9{{x}^{2}}} \right)$
$ \displaystyle \begin{align}u&=3{{x}^{3}}\\du&=9{{x}^{2}}\,dx\\dx&=\frac{{du}}{{9{{x}^{2}}}}\end{align}$ $ \displaystyle \begin{align}\color{#800000}{{\int{{\ln 4\left( {{{4}^{{3{{x}^{3}}}}}} \right)}}\left( {9{{x}^{2}}} \right)dx}}&=\ln 4\int{{\left( {{{4}^{u}}} \right)}}\left( {\cancel{{9{{x}^{2}}}}} \right)\frac{{du}}{{\cancel{{9{{x}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,&=\cancel{{\ln 4}}\cdot \frac{1}{{\cancel{{\ln 4}}}}\cdot {{4}^{u}}+C={{4}^{{3{{x}^{3}}}}}+C\end{align}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {f{{{\left( x \right)}}^{{g\left( x \right)}}}} \right]$ When we have a variable both in the base and the exponent, or if the function is really complicated, it’s best to take ln of both sides to take derivative, and use implicit integration. Then substitute $ y$ back in at the end. | Find Derivative: $ \displaystyle y={{\left( {3x} \right)}^{x}}$ $ \displaystyle \begin{align}\ln y&=\ln {{\left( {3x} \right)}^{x}};\,\,\,\,\,\ln y=x\ln \left( {3x} \right)\\\frac{1}{y}\left( {\frac{{dy}}{{dx}}} \right)&=\cancel{x}\cdot \frac{{\cancel{3}}}{{\cancel{3}\cancel{x}}}+\ln \left( {3x} \right)\cdot 1\,\,\,\,\text{(product rule)}\\\,\frac{{dy}}{{dx}}&=\left[ {1+\ln \left( {3x} \right)} \right]\cdot y={{\left( {3x} \right)}^{x}}\left[ {1+\ln \left( {3x} \right)} \right]={{3}^{x}}{{x}^{x}}\left[ {1+\ln \left( {3x} \right)} \right]\end{align}$ | |
| Integrals of the Six Basic Trig Functions $ \displaystyle \int{{\sin u\,du=-\cos u\,\,+\,\,C\,\,\,\,\,\,\,\,}}\int{{\cos u\,du=\sin u\,+\,C}}$ $ \displaystyle \int{{\tan u\,du=-\ln \left| {\cos u} \right|\,\,+\,\,C\,}}\,\,\,\,\,\,\,\,\int{{\cot u\,du=\ln \left| {\sin u} \right|\,+\,C}}$ $ \displaystyle \int{{\sec u\,du=\ln \left| {\sec u+\tan u} \right|\,\,+\,\,C}}\,\,\,\,\,\,\,\int{{\csc u\,du=-\ln \left| {\csc u+\cot u} \right|\,+\,C}}$ | $ \begin{align}u&=4\theta \\du&=4\,d\theta \\d\theta &=\frac{{du}}{4}\end{align}$ $ \displaystyle \begin{align}\color{#800000}{{\int{{\csc \left( {4\theta } \right)}}\,d\theta }}&=\int{{\csc u}}\,d\theta =\int{{\csc u\cdot \,\frac{{du}}{4}}}=\frac{1}{4}\int{{\csc u\,du}}\\&=-\frac{1}{4}\ln \left| {\csc u+\cot u} \right|\,+\,C\\&=-\frac{1}{4}\ln \left| {\csc \left( {4x} \right)+\cot \left( {4x} \right)} \right|\,+\,C\end{align}$ | |
Differentiation of the Natural Logarithmic
We’re jumping ahead here, but it’s interesting to note: when we learn the Power Rule for Integration here in the Antiderivatives and Integration section, we will notice that if $ n=-1$, the rule doesn’t apply, since the denominator would be $ 0$: $ \displaystyle \int{{{{x}^{n}}}}dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}\,+C,\,\,n\ne 1$. Thus, when we try to integrate a function like $ \displaystyle f\left( x \right)=\frac{1}{x}={{x}^{{-1}}}$, we have to do something “special”; namely learn that this integral is $ \ln \left( x \right)$.
Here are some logarithmic properties that we learned here in the Logarithmic Functions section; note we could use $ {{\log }_{a}}x$ instead of $ \ln x$. Remember that $ \ln x$ is the same as $ {{\log }_{e}}x$, where $ e\approx 2.718$ (“$ e$” is Euler’s Number). A log is the exponent raised to the base power ($ a$) to get the argument ($ x$) of the log; if “$ a$” is missing, we assume it’s 10.
Logarithmic Properties
$ \begin{align}\ln \left( {{{a}^{n}}} \right)=n\ln a\\\ln \left( {ab} \right)=\ln \left( a \right)+\ln \left( b \right)\\\ln \left( {\frac{a}{b}} \right)=\ln \left( a \right)-\ln \left( b \right)\\\ln \left( 1 \right)=0\\\ln \left( {{{e}^{x}}} \right)=x\end{align}$ $ \displaystyle \begin{align}&\text{Example of Expanding Log:}\\\ln {{\left( {\frac{{{{x}^{4}}y}}{{\sqrt{z}}}} \right)}^{2}}&=2\ln \left( {\frac{{{{x}^{4}}y}}{{{{z}^{{\frac{1}{2}}}}}}} \right)=2\left[ {\ln \left( {{{x}^{4}}} \right)+\ln \left( y \right)-\ln \left( {{{z}^{{\frac{1}{2}}}}} \right)} \right]\\&=2\left[ {4\ln \left( x \right)+\ln \left( y \right)-\frac{1}{2}\ln \left( z \right)} \right]\\&=8\ln \left( x \right)+2\ln \left( y \right)-\ln \left( z \right)\end{align}$
Natural Log Differentiation Rules:
Here are derivatives of natural logarithm functions:
| Natural Log (Ln) Differentiation Formula | Examples |
| $ \displaystyle \frac{d}{{dx}}\left( {\ln x} \right)=\frac{1}{x},\,\,\,\,x>0$ | $ \displaystyle \frac{d}{{dx}}\left( {5\ln x} \right)=5\cdot \frac{{d\left( {\ln x} \right)}}{{dx}}=5\cdot \frac{1}{x}=\frac{5}{x}$ |
| $ \displaystyle \frac{d}{{dx}}\left( {\ln u} \right)=\frac{1}{u}\times \frac{{du}}{{dx}}=\frac{{{u}’}}{u},\,\,\,\,u>0$ | $ \displaystyle \frac{d}{{dx}}\left[ {\ln \left( {2x} \right)} \right]=\frac{{{u}’}}{u}=\frac{{\frac{d}{{dx}}\left( {2x} \right)}}{{2x}}=\frac{2}{{2x}}=\frac{1}{x}$ |
Here are some natural log (ln) differentiation problems. Note that it’s typically easier to use the log properties to expand the function before differentiating. Also note that you may not have to simplify the answers as much as shown.
| Natural Logarithm Differentiation Problem | Solution |
| Find the derivative of the function:
$ \displaystyle f\left( x \right)=\ln \sqrt{{3{{x}^{2}}+2}}$ | $ \begin{align}f\left( x \right)&=\ln {{\left( {3{{x}^{2}}+2} \right)}^{{\frac{1}{2}}}}=\frac{1}{2}\ln \left( {3{{x}^{2}}+2} \right)\\{f}’\left( x \right)&=\frac{{{u}’}}{u}=\frac{1}{2}\left( {\frac{{6x}}{{3{{x}^{2}}+2}}} \right)=\frac{{3x}}{{3{{x}^{2}}+2}}\end{align}$ |
| Find the derivative of the function:
$ \displaystyle f\left( x \right)=\ln \left( {{{x}^{2}}\sqrt[3]{{\frac{{x+1}}{{x-1}}}}} \right)$ | $ \begin{align}f\left( x \right)&=\ln \left( {{{x}^{2}}\sqrt[3]{{\frac{{x+1}}{{x-1}}}}} \right)=2\ln \left( x \right)+\frac{1}{3}\ln \left( {x+1} \right)-\frac{1}{3}\ln \left( {x-1} \right)\,\,\,\,\text{(expand first)}\\{f}’\left( x \right)&=\frac{{{u}’}}{u}=2\cdot \frac{1}{x}+\frac{1}{3}\left( {\frac{1}{{x+1}}} \right)-\frac{1}{3}\left( {\frac{1}{{x-1}}} \right)=\frac{2}{x}+\frac{1}{{3\left( {x+1} \right)}}-\frac{1}{{3\left( {x-1} \right)}}\\&=\frac{{2\cdot 3\left( {{{x}^{2}}-1} \right)+x\left( {x-1} \right)-x\left( {x+1} \right)}}{{3x\left( {{{x}^{2}}-1} \right)}}=\frac{{6{{x}^{2}}-6+{{x}^{2}}-x-{{x}^{2}}-x}}{{3x\left( {{{x}^{2}}-1} \right)}}=\frac{{6{{x}^{2}}-2x-6}}{{3x\left( {{{x}^{2}}-1} \right)}}\end{align}$ |
| Find the derivative of the function:
$ \displaystyle y=\ln \left| {\frac{{\sin x}}{{\sin x-1}}} \right|$ | $ \require {cancel} \begin{align}y&=\ln \left| {\frac{{\sin x}}{{\sin x-1}}} \right|=\ln \left| {\sin x} \right|-\ln \left| {\sin x-1} \right|\\\frac{{dy}}{{dx}}&=\frac{{{u}’}}{u}=\frac{{\cos x}}{{\sin x}}-\frac{{\cos x}}{{\sin x-1}}=\cot x-\frac{{\cos x}}{{\sin x-1}}=\frac{{\cot x\left( {\sin x-1} \right)-\cos x}}{{\sin x-1}}\\&=\frac{{\cot x\cdot \sin x-\cot x-\cos x}}{{\sin x-1}}=\frac{{\frac{{\cos x}}{{\cancel{{\sin x}}}}\cdot \cancel{{\sin x}}-\cot x-\cos x}}{{\sin x-1}}=\frac{{-\cot x}}{{\sin x-1}}\end{align}$ |
| Find an equation of the tangent line at the point $ \left( {1,0} \right)$:
$ \displaystyle y={{x}^{2}}\ln x$ | $ \displaystyle \frac{{dy}}{{dx}}={{x}^{2}}\cdot \frac{1}{x}+\ln \left( x \right)\cdot 2x=x+2x\ln \left( x \right)$ (product rule) When $ x=1$, $ \displaystyle \frac{{dy}}{{dx}}=1+2\cdot 1\cdot \ln \left( 1 \right)=1+0=1$ Tangent line through point $ \left( {1,0} \right)$: $ y-0=1\left( {x-1} \right)$ or $ y=x-1$. |
| Use implicit integration to find $ \displaystyle \frac{{dy}}{{dx}}$:
$ 2{{x}^{3}}+\ln {{y}^{3}}+y=3x$ | $ \displaystyle 2{{x}^{3}}+\ln {{y}^{3}}+y=3x;\,\,\,\,\,\,6{{x}^{2}}+\frac{{3{{y}^{2}}}}{{{{y}^{3}}}}{y}’+{y}’=3$ $ \displaystyle {y}’\left( {\frac{3}{y}+1} \right)=3-6{{x}^{2}};\,\,\,\,\,\,{y}’=\frac{{3-6{{x}^{2}}}}{{\frac{3}{y}+1}}=\frac{{3-6{{x}^{2}}}}{{\frac{{3+y}}{y}}}=\frac{{y\left( {3-6{{x}^{2}}} \right)}}{{3+y}}=\frac{{3y\left( {1-2{{x}^{2}}} \right)}}{{3+y}}$ |
General Logarithmic Differentiation
Logarithmic Differentiation gets a little trickier when we’re not dealing with natural logarithms. Remember that from the change of base formula (for base $ a$) that $ \displaystyle {{\log }_{a}}x=\frac{{\log x}}{{\log a}}=\frac{{\ln x}}{{\ln a}}=\frac{1}{{\ln a}}\cdot \ln x$. The derivative is $ \displaystyle \frac{{d\left( {{{{\log }}_{a}}x} \right)}}{{dx}}=\frac{d}{{dx}}\left( {\frac{1}{{\ln a}}\cdot \ln x} \right)=\frac{1}{{\ln a}}\cdot \frac{d}{{dx}}\left( {\ln x} \right)=\frac{1}{{\ln a}}\cdot \frac{1}{x}=\frac{1}{{x\left( {\ln a} \right)}}$.We can also get the derivative of the exponential function $ y={{a}^{x}}$ using Implicit Differentiation:
Derivation of Derivative of $ \boldsymbol {{{a}^{x}}}$
$ \displaystyle y={{a}^{x}};\,\,\,\ln y=\ln {{a}^{x}}$
$ \displaystyle \ln y=x\ln a;\,\,\frac{{d\left( {\ln y} \right)}}{{dx}}=\frac{{d\left( {x\ln a} \right)}}{{dx}}$
$ \displaystyle \frac{1}{y}\frac{{dy}}{{dx}}=\ln a;\,\,\,\frac{{dy}}{{dx}}=\ln a\cdot y=y\ln a={{a}^{x}}\left( {\ln a} \right)$
Based on these derivations, here are the formulas for the derivative of the exponent and log functions:
| Exponential/Logarithmic Differentiation Formulas | Examples |
| $ \displaystyle \frac{d}{{dx}}\left( {{{{\log }}_{a}}x} \right)=\frac{1}{{x\left( {\ln a} \right)}},\,\,\,a\ne 1$ | $ \displaystyle \frac{d}{{dx}}\left( {{{{\log }}_{2}}x} \right)=\frac{1}{{x\left( {\ln 2} \right)}}$ |
| $ \displaystyle \frac{d}{{dx}}\left( {{{{\log }}_{a}}u} \right)=\frac{1}{{u\ln a}}\cdot \frac{{du}}{{dx}}=\frac{{{u}’}}{{u\left( {\ln \,a} \right)}}$ $ a\ne 1$ | $ \displaystyle \begin{align}\frac{d}{{dx}}\left[ {\log \left( {\sqrt{{4x}}} \right)} \right]&=\frac{d}{{dx}}\left[ {\log {{{\left( {4x} \right)}}^{{\frac{1}{2}}}}} \right]=\frac{d}{{dx}}\left[ {\frac{1}{2}\log \left( {4x} \right)} \right]\\\,&=\frac{1}{2}\frac{d}{{dx}}\left[ {\log \left( {4x} \right)} \right]=\frac{1}{2}\cdot \frac{4}{{\left( {4x} \right)\ln 10}}=\frac{1}{{2x\left( {\ln 10} \right)}}\end{align}$ $ \displaystyle (\text{also equals }\frac{1}{{x\cdot 2\left( {\ln 10} \right)}}=\frac{1}{{x\left( {\ln {{{10}}^{2}}} \right)}}=\frac{1}{{x\left( {\ln 100} \right)}})$ |
| $ \displaystyle \frac{d}{{dx}}\left( {{{a}^{x}}} \right)=\left( {\ln \,a} \right){{a}^{x}},\,\,a\ne 1$ | $ \displaystyle \frac{d}{{dx}}\left( {{{6}^{x}}} \right)=\left( {\ln 6} \right){{6}^{x}}$ (see above for implicit differentiation derivative example) |
| $ \displaystyle \frac{d}{{dx}}\left( {{{a}^{u}}} \right)=\left( {\ln \,a} \right){{a}^{u}}{u}’,\,\,\,a\ne 1$ | $ \displaystyle \frac{d}{{dx}}\left( {{{4}^{{3{{x}^{3}}}}}} \right)=\ln 4\left( {{{4}^{{3{{x}^{3}}}}}} \right)\left( {9{{x}^{2}}} \right)$ Alternate method, not using formula, but using implicit differentiation, and then substituting $ y$ back in: $ \displaystyle y={{4}^{{3{{x}^{3}}}}};\,\,\,\ln y=\ln \left( {{{4}^{{3{{x}^{3}}}}}} \right);\,\,\,\ln y=3{{x}^{3}}\ln 4$ $ \displaystyle \frac{{d\left( {\ln y} \right)}}{{dx}}=\frac{{d\left( {3{{x}^{3}}\ln 4} \right)}}{{dx}};\,\,\,\,\frac{1}{y}{y}’=\left( {\ln 4} \right)\left( {9{{x}^{2}}} \right)$ $ \displaystyle {y}’=\left( {\ln 4} \right)\left( {9{{x}^{2}}} \right)\cdot y=\left( {\ln 4} \right)\left( {9{{x}^{2}}} \right)\left( {{{4}^{{3{{x}^{3}}}}}} \right)=9\ln 4\left( {{{{64}}^{{{{x}^{3}}}}}} \right)\left( {{{x}^{2}}} \right)$ |
Note that for the last two problems above (exponential differentiation), we can just take the ln of each side and not worry about the “formula”. In fact, when we have a variable such as $ x$ in the base and also the exponent (such as $ y=f{{\left( x \right)}^{{g\left( x \right)}}}$), we need to take ln of both sides and use implicit differentiation to solve (called “logarithmic differentiation”). (Note: We can also use the method of taking the ln on both sides for differentiating complicated problems without logarithmic or exponential functions, such as $ \displaystyle y=\frac{{{{{\left( {2x+1} \right)}}^{3}}}}{{{{x}^{5}}\sqrt{{x+1}}}}$. See below for an example.)
Here’s an example (using product rule):
Differentiate $ y={{x}^{{x-1}}}$
$ \displaystyle \ln y=\ln \left( {{{x}^{{x-1}}}} \right);\,\,\,\,\,\ln y=\left( {x-1} \right)\ln x$
$ \displaystyle \frac{{d\left( {\ln y} \right)}}{{dx}}=\frac{{d\left[ {\left( {x-1} \right)\ln x} \right]}}{{dx}}$
$ \displaystyle \frac{1}{y}{y}’=\left( {x-1} \right)\cdot \frac{1}{x}+\ln x\cdot 1$
$ \displaystyle \begin{align}{y}’&=\left( {\frac{{x-1}}{x}+\ln x} \right)\cdot y={{x}^{{x-1}}}\left( {\frac{{x-1}}{x}+\ln x} \right)\\&={{x}^{{x-1}}}\left( {1-{{x}^{{-1}}}+\ln x} \right)={{x}^{{x-1}}}-{{x}^{{x-2}}}+{{x}^{{x-1}}}\ln x\end{align}$
Here are more logarithmic differentiation problems; note that typically want to expand logs before we integrate:
| Logarithmic Differentiation Problem | Solution |
| Find the derivative of the function:
$ \displaystyle f\left( x \right)=\log \left( {7x-2} \right)$ | $ \displaystyle \begin{align}\frac{{d\left( {{{{\log }}_{a}}u} \right)}}{{dx}}&=\frac{{{u}’}}{{u\ln a}}\\\frac{{d\left[ {{{{\log }}_{{10}}}\left( {7x-2} \right)} \right]}}{{dx}}&=\frac{7}{{\left( {7x-2} \right)\ln 10}}\end{align}$ |
| Find the derivative of the function:
$ \displaystyle f\left( x \right)=\log \left( {{{x}^{2}}\sqrt[3]{{\frac{{2x+1}}{{{{x}^{2}}-1}}}}} \right)$ | Expand first: $ \displaystyle f\left( x \right)=\log \left( {{{x}^{2}}\sqrt[3]{{\frac{{2x+1}}{{{{x}^{2}}-1}}}}} \right)=2\log \left( x \right)+\frac{1}{3}\log \left( {2x+1} \right)-\frac{1}{3}\log \left( {{{x}^{2}}-1} \right)$
$ \displaystyle \begin{align}{f}’\left( {{{{\log }}_{a}}u} \right)&=\frac{{{u}’}}{{u\ln a}}=\frac{1}{{\ln 10}}\left[ {2\cdot \frac{1}{x}+\frac{1}{3}\left( {\frac{2}{{2x+1}}} \right)-\frac{1}{3}\left( {\frac{{2x}}{{{{x}^{2}}-1}}} \right)} \right]\\&=\frac{1}{{\ln 10}}\left[ {\frac{2}{x}+\frac{2}{{3\left( {2x+1} \right)}}-\frac{{2x}}{{3\left( {{{x}^{2}}-1} \right)}}} \right]\end{align}$ (I could simplify with a common denominator, but I’m lazy 🙂 ). |
| Find the derivative of the function:
$ \displaystyle y=\frac{{5{{{\log }}_{2}}t}}{t}$ | Quotient Rule: $ \require{cancel} \begin{align}{y}’&=5\left[ {\frac{{t\cdot \frac{{d\left( {{{{\log }}_{2}}t} \right)}}{{dt}}-{{{\log }}_{2}}t\cdot 1}}{{{{t}^{2}}}}} \right]=\frac{5}{{{{t}^{2}}}}\left( {\cancel{t}\cdot \frac{1}{{\cancel{t}\ln 2}}-{{{\log }}_{2}}t} \right)\\&=\frac{5}{{{{t}^{2}}}}\left( {\frac{1}{{\ln 2}}-\frac{{\ln t}}{{\ln 2}}} \right)=\frac{5}{{{{t}^{2}}\ln 2}}\left( {1-\ln t} \right)\end{align}$ (I turned log into ln using Change of Base to simplify answer.) |
| Find an equation of the tangent line at the point $ \left( {3,1.5} \right)$:
$ \displaystyle y={{\log }_{2}}\sqrt{{{{x}^{2}}-1}}$ | $ \begin{align}y&={{\log }_{2}}{{\left( {{{x}^{2}}-1} \right)}^{{\frac{1}{2}}}}=\frac{1}{2}{{\log }_{2}}\left( {{{x}^{2}}-1} \right)\\{y}’&=\frac{{{u}’}}{{u\ln a}}=\frac{1}{2}\left[ {\frac{{2x}}{{\left( {{{x}^{2}}-1} \right)\ln 2}}} \right]=\frac{x}{{\left( {{{x}^{2}}-1} \right)\ln 2}}\end{align}$ When $ x=3$, $ \displaystyle {y}’=\frac{3}{{\left( {{{3}^{2}}-1} \right)\ln \left( 2 \right)}}\approx .541$. Tangent line through point $ \left( {3,1.5} \right)$: $ y-1.5=.541\left( {x-3} \right)$ or $ y=.541x-.123$. |
Here are more problems where we take the ln of both sides. Note that the first problem isn’t even a log or exponent problem, but we’ll take the ln of both sides to make it much easier to differentiate!
| Differentiation Problem | Solution |
| Find the derivative of the function using logarithmic differentiation:
$ \displaystyle y=\frac{{{{{\left( {2x+1} \right)}}^{3}}}}{{{{x}^{5}}\sqrt{{x+1}}}}$ | Even though this isn’t a logarithmic or exponential problem, it’s a difficult problem to differentiate, so we can take the ln of each side to make it easier: $ \displaystyle y=\frac{{{{{\left( {2x+1} \right)}}^{3}}}}{{{{x}^{5}}\sqrt{{x+1}}}};\,\,\,\ln y=\ln \left( {\frac{{{{{\left( {2x+1} \right)}}^{3}}}}{{{{x}^{5}}\sqrt{{x+1}}}}} \right)$ $ \displaystyle \begin{align}\ln y&=3\ln \left( {2x+1} \right)-5\ln x-\frac{1}{2}\ln \left( {x+1} \right)\\\frac{1}{y}\left( {\frac{{dy}}{{dx}}} \right)&=\frac{6}{{2x+1}}-\frac{5}{x}-\frac{1}{{2\left( {x+1} \right)}}\\\frac{{dy}}{{dx}}&=\left( {\frac{6}{{2x+1}}-\frac{5}{x}-\frac{1}{{2x+2}}} \right)\cdot y\\&=\left( {\frac{6}{{2x+1}}-\frac{5}{x}-\frac{1}{{2x+2}}} \right)\cdot \frac{{{{{\left( {2x+1} \right)}}^{3}}}}{{{{x}^{5}}\sqrt{{x+1}}}}\end{align}$ (Yes, we could probably simplify, but we’ll decide not to ?) |
| Find the derivative of the function, by using logarithmic differentiation:
$ \displaystyle y={{x}^{{3x-1}}}$ | $ \displaystyle y={{x}^{{3x-1}}};\,\,\,\,\ln y=\ln {{x}^{{3x-1}}};\,\,\,\,\ln y=\left( {3x-1} \right)\ln x$ $ \displaystyle \begin{align}\frac{1}{y}\left( {\frac{{dy}}{{dx}}} \right)&=\left( {3x-1} \right)\frac{1}{x}+\ln x\cdot 3\,\,\text{(product rule)}\\\frac{{dy}}{{dx}}&=\left[ {\left( {3x-1} \right)\frac{1}{x}+3\ln x} \right]\cdot y\\&={{x}^{{3x-1}}}\left[ {\frac{{3x-1}}{x}+3\ln x} \right]=3{{x}^{{3x-1}}}-{{x}^{{3x-2}}}+3{{x}^{{3x-1}}}\ln x\end{align}$ |
| Find the derivative of the function, by using logarithmic differentiation:
$ f\left( x \right)={{\left( {3x+4} \right)}^{{{{x}^{2}}}}}$ | $ \displaystyle y={{\left( {3x+4} \right)}^{{{{x}^{2}}}}};\,\,\,\,\,\ln y=\ln {{\left( {3x+4} \right)}^{{{{x}^{2}}}}};\,\,\,\,\ln y={{x}^{2}}\ln \left( {3x+4} \right)$ $ \displaystyle \begin{align}\frac{1}{y}\left( {\frac{{dy}}{{dx}}} \right)&={{x}^{2}}\cdot \frac{3}{{3x+4}}+\ln \left( {3x+4} \right)\cdot 2x\,\,\,\,\text{(product rule)}\\\frac{{dy}}{{dx}}&=\left[ {\frac{{3{{x}^{2}}}}{{3x+4}}+2x\cdot \ln \left( {3x+4} \right)} \right]\cdot y={{\left( {3x+4} \right)}^{{{{x}^{2}}}}}\left[ {\frac{{3{{x}^{2}}}}{{3x+4}}+2x\ln \left( {3x+4} \right)} \right]\\&=x{{\left( {3x+4} \right)}^{{{{x}^{2}}}}}\left[ {\frac{{3x}}{{3x+4}}+2\ln \left( {3x+4} \right)} \right]\end{align}$ |
| Find the derivative of the function (don’t have to use logarithmic differentiation since the base isn’t a variable):
$ \displaystyle g\left( \theta \right)=\frac{{{{4}^{{\theta -2}}}}}{{\sin \left( {\pi \theta } \right)}}$ | $ \displaystyle \begin{align}{g}’\left( \theta \right)&=\frac{{\sin \left( {\pi \theta } \right)\cdot \ln 4\cdot {{4}^{{\theta -2}}}-{{4}^{{\theta -2}}}\cdot \cos \left( {\pi \theta } \right)\cdot \pi }}{{{{{\sin }}^{2}}\left( {\pi \theta } \right)}}\,\,\,\,\,\,\,\text{(quotient rule)}\\&=\frac{{{{4}^{{\theta -2}}}\left[ {\ln 4\cdot \sin \left( {\pi \theta } \right)-\pi \cos \left( {\pi \theta } \right)} \right]}}{{{{{\sin }}^{2}}\left( {\pi \theta } \right)}}={{4}^{{\theta -2}}}\left[ {\frac{{\ln 4\cdot \sin \left( {\pi \theta } \right)}}{{{{{\sin }}^{2}}\left( {\pi \theta } \right)}}-\frac{{\pi \cos \left( {\pi \theta } \right)}}{{{{{\sin }}^{2}}\left( {\pi \theta } \right)}}} \right]\\&={{4}^{{\theta -2}}}\left[ {\ln 4\cdot \csc \left( {\pi \theta } \right)-\pi \cot \left( {\pi \theta } \right)\csc \left( {\pi \theta } \right)} \right]\end{align}$ |
Here’s one more that uses Implicit Differentiation:
| Exponential Differentiation Problem | Solution |
| Find $ {y}’$ by implicit differentiation:
$ \displaystyle {{y}^{x}}=\ln x$ | Take the ln of each side, even though we’ll have the ln of an ln: $ {{y}^{x}}=\ln x;\,\,\,\,\ln {{y}^{x}}=\ln \left( {\ln x} \right);\,\,\,x\ln y=\ln \left( {\ln x} \right)$ Now, differentiate and solve for $ {y}’$: $ \begin{align}x\ln y&=\ln \left( {\ln x} \right)\\x\cdot \frac{1}{y}{y}’+\ln y\cdot 1&=\frac{{\frac{1}{x}}}{{\ln x}}=\frac{1}{{x\ln x}}\\\frac{x}{y}\cdot {y}’+\ln y&=\frac{1}{{x\ln x}};\,\,\,\,\frac{x}{y}\cdot {y}’=\frac{1}{{x\ln x}}-\ln y\\{y}’&=\left( {\frac{1}{{x\ln x}}-\ln y} \right)\left( {\frac{y}{x}} \right)\end{align}$ This can also be written as $ \displaystyle \left( {\frac{1}{{x{{y}^{x}}}}-\ln y} \right)\left( {\frac{y}{x}} \right)=\frac{y}{{{{x}^{2}}{{y}^{x}}}}-\frac{{y\ln y}}{x}$. |
Derivative of eu
Yeah! This is actually the easiest function to differentiate, since $ \displaystyle \frac{d}{{dx}}\left( {{{e}^{x}}} \right)={{e}^{x}}$! I know; it’s strange, isn’t it? This means that the slope of the graph of this function at any point is just equal to the $ y$-coordinate of that point. When we have a function of $ x$ in the exponent, we just have to multiply by the derivative of this function: $ \displaystyle \frac{d}{{dx}}\left( {{{e}^{u}}} \right)={{e}^{u}}\frac{{du}}{{dx}}$.
Here are some problems:
| $ \boldsymbol {{{e}^{x}}}$ Differentiation Problem | Solution | $ \boldsymbol {{{e}^{x}}}$ Differentiation Problem | Solution |
| Find the derivative of
$ f\left( x \right)={{e}^{{3x}}}$ | $ {f}’\left( x \right)={{e}^{{3x}}}\cdot 3=3{{e}^{{3x}}}$ | Find the derivative of
$ y={{x}^{2}}{{e}^{x}}$ | $ \displaystyle {y}’={{x}^{2}}\cdot {{e}^{x}}+{{e}^{x}}\cdot 2x={{e}^{x}}x\left( {x+2} \right)$ |
| Find the derivative of
$ y=\ln \left( {1+2{{e}^{{3x}}}} \right)$ | $ \displaystyle {y}’=\frac{{{u}’}}{u}=\frac{{6{{e}^{{3x}}}}}{{1+2{{e}^{{3x}}}}}$ | Find the derivative of
$ \displaystyle f\left( x \right)=\frac{{{{e}^{{-x}}}}}{{1+{{e}^{{-x}}}}}$ | $ \require {cancel} \displaystyle \begin{align}{f}’\left( x \right)&=\frac{{\left( {1+{{e}^{{-x}}}} \right)\left( {-{{e}^{{-x}}}} \right)-\left( {{{e}^{{-x}}}} \right)\left( {-{{e}^{{-x}}}} \right)}}{{{{{\left( {1+{{e}^{{-x}}}} \right)}}^{2}}}}\\&=\frac{{-{{e}^{{-x}}}\cancel{{-{{e}^{{-2x}}}}}\cancel{{+{{e}^{{-2x}}}}}}}{{{{{\left( {1+{{e}^{{-x}}}} \right)}}^{2}}}}=-\frac{{{{e}^{{-x}}}}}{{{{{\left( {1+{{e}^{{-x}}}} \right)}}^{2}}}}\,\,\,\,(=-\frac{{{{e}^{x}}}}{{{{{\left( {1+{{e}^{x}}} \right)}}^{2}}}})\end{align}$ |
| Find the equation of the tangent line at
$ x{{e}^{x}}-{{e}^{{x-y}}}=0$
at point $ \left( {1,0} \right)$ | First use implicit differentiation to solve for $ {y}’$:
$ \displaystyle \begin{array}{c}x{{e}^{x}}-{{e}^{{x-y}}}=0\\\left( {x{{e}^{x}}+{{e}^{x}}\cdot 1} \right)-{{e}^{{x-y}}}\left( {1-{y}’} \right)=0\\x{{e}^{x}}+{{e}^{x}}-{{e}^{{x-y}}}+{{e}^{{x-y}}}{y}’=0\\{{e}^{{x-y}}}{y}’={{e}^{{x-y}}}-x{{e}^{x}}-{{e}^{x}}\end{array}$ $ \displaystyle {y}’=\frac{{{{e}^{{x-y}}}-x{{e}^{x}}-{{e}^{x}}}}{{{{e}^{{x-y}}}}}=1-x{{e}^{y}}-{{e}^{y}}$
Tangent Line: $ \begin{array}{l}{f}’\left( {1,0} \right)=1-\left( 1 \right){{e}^{0}}-{{e}^{0}}=-1\\y-0=-1\left( {x-1} \right);\,\,\,\,y=-x+1\end{array}$
| Find the second derivative of
$ \left( {x+2} \right){{e}^{{-2x}}}$ | $ \begin{align}{y}’&=\left( {x+2} \right)\cdot -2{{e}^{{-2x}}}+{{e}^{{-2x}}}\cdot 1\\&=-2x{{e}^{{-2x}}}-4{{e}^{{-2x}}}+{{e}^{{-2x}}}\\&=-2x{{e}^{{-2x}}}-3{{e}^{{-2x}}}=-{{e}^{{-2x}}}\left( {2x+3} \right)\end{align}$
$ \begin{align}{y}^{\prime \prime}&=-{{e}^{{-2x}}}\cdot 2+\left( {2x+3} \right)\cdot 2{{e}^{{-2x}}}\\&=-2{{e}^{{-2x}}}+4x{{e}^{{-2x}}}+6{{e}^{{-2x}}}\\&=4x{{e}^{{-2x}}}+4{{e}^{{-2x}}}=4{{e}^{{-2x}}}\left( {x+1} \right)\end{align}$ |
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