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Log and Exponential Differentiation Exercises Worked

Calculate the derivative of each of the following functions. When possible, simplify the given expression first.

[Textbook problems may use {$ \log_{10} $} for base 10 logarithms and just {$ \log \text{ or } \log_e $} for natural logarithms. Solutions worked use {$ \log $} for base 10 logs and {$ \ln $} for natural logarithms. ]

C65-1. {$$ \log_{10} x^3 $$}

Simplify first:

{$$ \log x^3 = 3 log x $$}

{$$ \begin{align} {d \over {dx}} 3 \log x &= 3{d \over {dx}} \log x \cr &= {{3\log e} \over x }\end{align} $$}


C65-1. {$ \color{blue}{ y =\log_{10} x^3} \text{ and } \color{red}{y^\prime} $}

C65-2. {$ \color{blue}{ y =\log_{10} \sqrt{x}} \text{ and } \color{red}{y^\prime} $}

C65-2. {$$ \log_{10} \sqrt{x} $$}

Simplify first:

{$$ \log \sqrt x = \log x^{1 \over 2} = {1 \over 2} \log x $$}

{$$ \begin{align} {d \over {dx}}{1 \over 2} \log x &= {1 \over 2}{d \over {dx}}\log x \cr &= ({1 \over 2}) {{\log e} \over x} \cr &= {{\log e} \over {2x}} \end{align} $$}

Which comes to the same thing as working the original by the chain rule:

{$$ \begin{align} {{d \over {dx}} \log_{10} \sqrt{x}} &= {{1 \over {\sqrt{x}}}(\log_{10}e){1 \over 2}x^{-{1 \over 2}}}\cr &= {1 \over 2} {1 \over {\sqrt{x}}} {1 \over {\sqrt{x}}}(\log_{10}e) \cr &= {{\log_{10}e} \over {2x}}\end{align} $$}


C65-3. {$ \color{blue}{ y =\log_{10}(1+x)} \text{ and } \color{red}{y^\prime} $}

C65-3. {$$ \log_{10} (1+2x) $$}

{$$ \begin{align} {d \over {dx}} \log_{10} (1+2x) &= {1 \over {1+x}}(\log_{10}e)2 \cr &= {{2\log_{10}e} \over {1+x}}\end{align} $$}


C65-4. {$ \color{blue}{ y =\log_{10}(1+x^3)} \text{ and } \color{red}{y^\prime} $}

C65-4. {$$ \log_{10} (1+x^3) $$}

{$$ \begin{align} {d \over {dx}}\log_{10} (1+x^3) &= {1 \over {1+x^3}}(\log_{10}e)3x^2 \cr &= {{3x^2\log_{10}e} \over {1+x^3}}\end{align} $$}


C65-5. {$ \color{blue}{ y =\ln(1+x)^3} \text{ and } \color{red}{y^\prime} $}

C65-5. {$$ \log_e (1+x)^3 $$}

Simplify first, per instructions.

{$$ \begin{align} {d \over {dx}}\ln (1+x)^3 &= {d \over {dx}}3 \ln (1+x) \cr &= 3{d \over {dx}} \ln (1+x) \cr &= 3 ({1 \over {1+x}}) \cr &= {3 \over {1+x}} \end{align} $$}


C65-6. {$ \color{blue}{ y =\ln\sqrt{3+5x}} \text{ and } \color{red}{y^\prime} $}

C65-6. {$$ \log_e \sqrt{3+5x} $$}

{$$ \begin{align}{d \over {dx}}\ln \sqrt{3+5x} &= {1 \over 2} {d \over {dx}} \ln (3+5x) \cr &= {1 \over 2}{1 \over {3+5x}}5 \cr &= {5 \over {6+ 10x}}\end{align} $$}


C65-7. {$ \color{blue}{ y =\ln {1 \over x}} \text{ and } \color{red}{y^\prime} $}

C65-7. {$$ \log_e ({1 \over x})$$}

{$$ \begin{align} {d \over {dx}} \ln ({1 \over x}) &= -1 {d \over {dx}} \ln x \cr &= (-1){1 \over x} \cr &= - {1 \over x} \end{align} $$}


C65-8. {$ \color{blue}{ y =\log x^{-3}} \text{ and } \color{red}{y^\prime} $}

C65-8. {$$ \log_{10} (x^{-3}) $$}

{$$ \begin{align} {d \over {dx}} \log (x^{-3}) &= -3 {d \over {dx}} \log x \cr &= -3{ {\log e} \over x} \cr &= - {{3\log e} \over x} \end{align} $$}


C65-9. {$ \color{blue}{ y = x \ln x^2} \text{ and } \color{red}{y^\prime} $}

C65-9. {$$ x \log_e x^2 $$}

{$$ \begin{align} {d \over {dx}} x \ln x^2 &= {d \over {dx}} 2x \ln x \cr &= 2x{d \over {dx}}\ln x + (\ln x){d \over {dx}}2x \cr &= {{2x} \over x} + 2 \ln x \cr &= 2 + 2\ln x \end{align} $$}


C65-10. {$ \color{blue}{ y = \ln{{1-x} \over {1+x}}} \text{ and } \color{red}{y^\prime} $}

C65-10. {$$ \log_e \left( {{1-x} \over {1+x}} \right) $$}

Simplify: {$$ \begin{align} \ln \left( {{1-x} \over {1+x}} \right) &= \ln \left[ (1-x)(1+x)^{-1} \right] \cr &= \ln (1-x) - \ln (1+x) \end{align} $$}

{$$ \begin{align} {d \over {dx}}\left[ \ln (1-x) - \ln (1+x)\right] &= {d \over {dx}}\ln(1-x) - {d \over {dx}}\ln(1+x) \cr &= \left( {1 \over {1-x}} \right)(-1) - \left({1 \over {1+x}} \right)(1) \cr &= {{-(1+x) -(1-x)} \over (1-x)(1+x)} \cr &= -{2 \over {1-x^2}} \end{align} $$}


C65-11. {$ \color{blue}{ y = \ln(2 - {t \over {1-t}})} \text{ and } \color{red}{y^\prime} $}

C65-11. {$$ \log_{10} \left( 2 - {t \over {1-t}} \right) $$}

{$$ \begin{align} \log \left( 2 - {t \over{1-t}} \right) &= \log \left( {{2-2t-t} \over {1-t}} \right) \cr &= \log \left( {{2-3t} \over {1-t}} \right) \cr &= \log \left[ (2-3t)(1-t)^{-1} \right] \cr &= \log(2-3t) - \log(1-t) \cr \end{align} $$}

{$$ \begin{align} {d \over {dt}} \left[ \log(2-3t) - \log(1-t) \right] &= {d \over {dt}} \left[ \log(2-3t) \right] - {d \over {dt}} \left[ \log(1-x) \right] \cr &= {{\log e} \over {2-3t}}(-3) - {{\log e} \over {1-t}}(-1) \cr &= {{\log e(-3+3t+2-3t)} \over (2-3t)(1-t)} \cr &= -{{\log e} \over (2-3t)(1-t)} \end{align} $$}

Warning! Does not agree in sign with textbook answer (but textbook answer is geometrically implausible).


C65-12. {$ \color{blue}{ y = \ln {t \over {\sqrt{1-t^2}}}} \text{ and } \color{red}{y^\prime} $}

C65-12. {$$ \log_e {t \over {\sqrt{1-t^2}}}$$}

{$$ \begin{align} \ln {t \over {\sqrt{1-t^2}}} &= \ln t - {1 \over 2} \ln(1-t^2) \cr {d \over {dt}}\left[ \ln t - {1 \over 2} \ln(1-t^2) \right] &= {d \over {dt}}(\ln t) - {1 \over 2}{d \over {td}}\ln(1-t^2) \cr &= {1 \over t} - {1 \over 2}({1 \over {1-t^2}})(-2t) \cr &= {1 \over t} + {t \over {1-t^2}} \cr &= {{1-t^2+t^2} \over {t(1-t^2)}} \cr {d \over {dt}}\left( \ln {t \over {\sqrt{1-t^2}}} \right) &= {1 \over {t(1=t^2)}}\end{align} $$}


C65-13. {$ \color{blue}{ y = {{\ln t} \over {t^2}}} \text{ and } \color{red}{y^\prime} $}

C65-13. {$$ {{\log_e t} \over t} $$}

{$$ \begin{align} {d \over {dt}}\left( {{\ln t} \over t}\right) &= {{t{d \over {dt}}\ln t - \ln t{d \over {dt}}t} \over {t^2}} \cr &= {{t({1 \over t}) - \ln t} \over {t^2}} \cr {d \over {dt}}\left( {{\ln t} \over t}\right) &= {{1- \ln t} \over {t^2}}\end{align} $$}


C65-14. {$ \color{blue}{ y = \ln \{ln x \}} \text{ and } \color{red}{y^\prime} $}

C65-14. {$$ \log_e \{log_e x \} $$}

{$$ \begin{align} {d \over {dx}}\left( \ln \{ln x \}\right) &= {1 \over {\ln x}}{1 \over x} \cr {d \over {dx}}\left( \ln \{ln x \}\right) &= {1 \over {x \ln x}} \end{align} $$}


C65-15. {$ \color{blue}{ y = (\ln t)^4} \text{ and } \color{red}{y^\prime} $}

C65-15. {$$ (\log_e t)^4 $$}

{$$ \begin{align} {d \over {dt}}(\ln t)^4 &= 4(\ln t)^3({1 \over t}) \cr {d \over {dt}}(\ln t)^4 &= {{4(\ln t)^3} \over t}\end{align} $$} Warning! Does not agree with textbook answer.


C67-1. {$ \color{blue}{ y = e^{3x}} \text{ and } \color{red}{y^\prime} $}

C67-1. {$$ e^{3x} $$}

{$$ {d \over {dx}}e^{3x} = 3e^{3x} $$}

The point of these is {$ {d \over {dx}} e^x = e^x $} and beyond that, most of these exercises are simply application of the Chain rule.


C67-2. {$ \color{blue}{ y = e^{2x+x^2}} \text{ and } \color{red}{y^\prime} $}

C67-2. {$$ e^{2x+x^2} $$}

{$$ \begin{align} {d \over {dx}}e^{2x+x^2} &= e^{2x+x^2}(2 + 2x) \end{align}$$}


C67-3. {$ \color{blue}{ y =e^{\sqrt{1+x}} } \text{ and } \color{red}{y^\prime} $}

C67-3. {$$ e^{\sqrt{1+x}} $$}

{$$ \begin{align} {d \over {dx}} e^{\sqrt{1+x}} &= e^{\sqrt{1+x}}({1 \over 2})(1+x)^{-{1 \over 2}} \cr &= e^{\sqrt{1+x}}\left( {1 \over {2\sqrt{1+x}}} \right)\cr {d \over {dx}} e^{\sqrt{1+x}} &={{e^{\sqrt{1+x}}} \over {2\sqrt{1+x}}} \end{align} $$}


C67-4. {$ \color{blue}{ y = e^{\ln x}} \text{ and } \color{red}{y^\prime} $}

C67-4. {$$ e^{\log x} $$}

{$$ \begin{align} e^{\ln x} &= x \cr {d \over {dx}}x &= 1 \end{align} $$}


C67-5. {$ \color{blue}{ y = x^2e^x} \text{ and } \color{red}{y^\prime} $}

C67-5. {$$ x^2e^x $$}

{$$ \begin{align} {d \over {dx}}\left( x^2e^x \right) &= e^x{dx \over {dx}}x^2 + x^2{d \over {dx}}e^x \cr &= e^x2x + x^2e^x \cr &= (2x + x^2)e^x \end{align} $$}


C67-6. {$ \color{blue}{ y = (1-x)^3e^{x^2}} \text{ and } \color{red}{y^\prime} $}

C67-6. {$$ (1-x)^3e^{x^2} $$}

{$$ \begin{align} {d \over {dx}}\left[ (1-x)^3e^{x^2} \right] &= e^{x^2}{d \over {dx}}(1-x)^3 \cr &+ (1-x)^3{d \over {dx}}e^{x^2} \cr &= e^{x^2}(3)(1-x)^2(-1) \cr &+ (1-x)^3(2x)e^{x^2} \end{align} $$}

{$$ \begin{align} {d \over {dx}}\left[ (1-x)^3e^{x^2} \right] &= e^{x^2}(1-x)^2 \left( 2x(1-x) - 3 \right) \cr &= e^{x^2}(1-x)^2 \left(2x- 2x^2 -3 \right) \end{align} $$}


C67-7. {$ \color{blue}{ y = 10^{3x+4}} \text{ and } \color{red}{y^\prime} $}

C67-7. {$$ 10^{3x+4} $$}

Solution proceeds by logarithmic differentiation.

{$$ \begin{align} \text{Let } u &= 10^{3x + 4} \cr (\ln u)^\prime &= {{u^\prime} \over u} \text{ (Chain rule)} \cr (\ln 10^{3x+4})^\prime &= {{u^\prime} \over {10^{3x+4}}} \cr ((3x+4)\ln 10)^\prime &= {{u^\prime} \over {10^{3x+4}}} \cr (\ln 10)(3x+4)^\prime &= {{u^\prime} \over {10^{3x+4}}} \cr 3 \ln 10 (10^{3x+4}) &= u^\prime \end{align} $$}


C67-8. {$ \color{blue}{ y = a^{(1+x)^3}} \text{ and } \color{red}{y^\prime} $}

C67-8. {$$ a^{(1+x)^3} $$}

Solution proceeds by logarithmic differentiation.

{$$ \begin{align} \text{Let } u &= a^{(1+x)^3} \cr (\ln u)^\prime &= {{u^\prime} \over u} \cr \end{align} $$}

{$$ \begin{align} (\ln a^{(1+x)^3})^\prime &= {{u^\prime} \over {a^{(1+x)^3}}} \cr ((1+x)^3\ln a)^\prime &= \cr \ln a((1+x)^3)^\prime &= \cr \ln a(3(1+x)^2) &= {{u^\prime} \over {a^{(1+x)^3}}} \cr 3(1+x)^2(\ln a)a^{(x+1)^3} &= u^\prime \end{align} $$}


C67-9. {$ \color{blue}{ y = \ln e^x } \text{ and } \color{red}{y^\prime} $}

C67-9. {$$ \log e^x $$}

This is just exercise C67-4 the other way around. Since exponentiation and the natural log are defined as inverse functions, the natural logarithm of {$ e^x $} is just {$ x $}, and the derivative of x with respect to x is 1.


C67-10. {$ \color{blue}{ y = \ln (1+e^x)} \text{ and } \color{red}{y^\prime} $}

C67-10. {$$ \log (1+e^x) $$}

{$$ \begin{align} {d \over {dx}}\ln (1+e^x) &= {1 \over {1+e^x}} e^x \cr {d \over {dx}}\ln (1+e^x) &= {{e^x} \over {1+e^x}} \end{align} $$}


C67-11. {$ \color{blue}{ y = \ln e^{-x^2}} \text{ and } \color{red}{y^\prime} $}

C67-11. {$$ \log e^{-x^2} $$}

Simplify:

{$$ \ln e^{-x^2} = -x^2 $$}

{$$ {d \over {dx}}( -x^2) = -2x $$}


C67-12. {$ \color{blue}{ y = (\ln e^{2x})^2 } \text{ and } \color{red}{y^\prime} $}

C67-12. {$$ (\log e^{2x})^2 $$}

Simplify:

{$$ (\ln e^{2x})^2 = (2x)^2 = 4x^2 $$}

{$$ {d \over {dx}}4x^2 = 2(4x) = 8x $$}


C67-13. {$ \color{blue}{ y = (e^x + 1)^2 } \text{ and } \color{red}{y^\prime} $}

C67-13. {$$ (e^x + 1)^2 $$}

{$$ {d \over {dx}}(e^x +1)^2 = 2(e^x+1)e^x = 2e^x(e^x+1) $$}


C67-14. {$ \color{blue}{ y = {{e^{\sqrt{x}} + e^{-\sqrt{x}} } \over 2} } \text{ and } \color{red}{y^\prime} $}

C67-14. {$$ {{e^{\sqrt{x}} + e^{-\sqrt{x}} } \over 2} $$}

{$$ \begin{align} & {{e^{\sqrt{x}} + e^{-\sqrt{x}} } \over 2} \cr &= {1 \over 2}e^{x^{1 \over 2}} + {1 \over 2}e^{-x^{1 \over 2}} \cr {d \over {dx}}\left( {{e^{\sqrt{x}} + e^{-\sqrt{x}} } \over 2} \right) &= {d \over {dx}}\left( {1 \over 2}e^{x^{1 \over 2}} \right) + {d \over {dx}}\left( {1 \over 2}e^{-x^{1 \over 2}} \right) \end{align} $$}

{$$ \begin{align} {d \over {dx}}\left( {{e^{\sqrt{x}} + e^{-\sqrt{x}} } \over 2} \right) &= {1 \over 2}{d \over {dx}}\left( e^{x^{1 \over 2}} \right) + {1 \over 2}{d \over {dx}}\left( e^{-x^{1 \over 2}} \right) \cr &= {1 \over 2}( e^{x^{1 \over 2}}){d \over {dx}}\left( x^{1 \over 2} \right) + {1 \over 2}( e^{-x^{1 \over 2}} ){d \over {dx}}\left( -x^{1 \over 2} \right) \cr &= {1 \over 2}( e^{x^{1 \over 2}}){1 \over 2}( x^-{1 \over 2}) + {1 \over 2}( e^{-x^{1 \over 2}} )-{1 \over 2}( x^-{1 \over 2} ) \cr &= {1 \over {4\sqrt{x}}}( e^{x^{1 \over 2}}) - {1 \over {4\sqrt{x}}}( e^{-x^{1 \over 2}}) \cr &= {{ e^{\sqrt{x}}}\over {4\sqrt{x}}} - {1 \over {4\sqrt{x}(e^{\sqrt{x}})}} \end{align} $$}


C67-15. {$ \color{blue}{ y = {{e^{\sqrt{x}} - e^{-\sqrt{x}} } \over 2} } \text{ and } \color{red}{y^\prime} $}

C67-15. {$$ {{e^{\sqrt{x}} - e^{-\sqrt{x}} } \over 2} $$}

{$$ \begin{align} {d \over {dx}}\left ({{e^{\sqrt{x}} - e^{-\sqrt{x}} } \over 2} \right) &= {(e^{\sqrt{x}}{1 \over 2}x^{-{1 \over 2}} - e^{-\sqrt{x}}({{-1} \over 2})x^{-{1 \over 2}} ) \over 2} \cr &= {{e^{\sqrt{x}} + e^{-\sqrt{x}} } \over {4\sqrt{x}}} \end{align} $$}


C67-16. {$ \color{blue}{ y = {{e^x - e^{-x}} \over {e^x + e^{-x}}} } \text{ and } \color{red}{y^\prime} $}
aka {$ \color{blue}{ y = \tanh(x) } \text{ and } \color{red}{y= \operatorname{sech}^2(x)}$}

C67-16. {$$ {{e^x - e^{-x}} \over {e^x + e^{-x}}} $$}

This is the hyperbolic tangent {$ \tanh $}.

Since,

{$$ \begin{gather} \sinh x = {{e^x-e^{-x}} \over 2}, \text{ and } \cosh x = {{e^x-e^{-x}} \over 2} \cr \tanh(x) = {{{e^x-e^{-x}} \over 2} \over {{e^x-e^{-x}} \over 2}} = {{e^x - e^{-x}} \over {e^x + e^{-x}}} \end{gather} $$}

{$$ \begin{align} {d \over {dx}}\left( {{e^x - e^{-x}} \over {e^x + e^{-x}}}\right) &= {(e^x + e^{-x}){d \over {dx}}(e^x - e^{-x})-(e^x - e^{-x}){d \over {dx}}(e^x + e^{-x}) \over {(e^x + e^{-x})^2}} \cr &= {{(e^x + e^{-x})^2-(e^x - e^{-x})^2} \over {(e^x + e^{-x})^2}} \end{align} $$}

As with most trig-like functions, there are several ways to go from here:

For example:

{$$ \begin{align}{{(e^x + e^{-x})^2-(e^x - e^{-x})^2} \over {(e^x + e^{-x})^2}} &= 1 - {{(e^x - e^{-x})^2} \over {(e^x + e^{-x})^2}} = 1 - \tanh^2(x), \text{ or } \cr {{(e^x + e^{-x})^2-(e^x - e^{-x})^2} \over {(e^x + e^{-x})^2}} &= {{{{(e^x + e^{-x})^2} \over 4} -{{(e^x - e^{-x})^2} \over 4}} \over {{1 \over 4}(e^x + e^{-x})^2}} \cr &= {{\cosh^2(x) - \sinh^2(x) } \over { {1 \over 4} (e^x + e^{-x})^2}} \cr &= {1 \over { {1 \over 4} (e^x + e^{-x})^2}} = {4 \over {(e^x + e^{-x})^2}} \cr &= {1 \over \cosh^2(x) } = \operatorname{sech}^2(x) = 1 - \tanh^2(x) \end{align} $$}


Sources:

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Category: Math Calculus


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August 05, 2017

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