« Exponential Equations |
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1. $\ln1 = \ln{e^0} = 0$
2. $\ln{e} = \ln{e^1} = 1$
3. $\ln{\frac{1}{e}} = \ln{e^{-1}} = -1$
4. $\ln{e^x} = x$
5. $\ln{e^{ln \, x}} = x$ (for $x > 0$)
6. $b^x = e^{x \ln{b}}$ (for $b > 0$)
7. $\ln{xy} = \ln{x} + \ln{y}$
8. $\ln \left( {\frac{x}{y}} \right) = \ln x - \ln y$
9. $\ln {( x^n )} = n\ln x$
10. ${\log _a}x = \frac{\ln x}{\ln a}$ (for $a > 0$, $a \ne 1$, $x > 0$)
1. $\frac{d}{dx}\left[ {\ln x} \right] = \frac{1}{x}$ (for $x > 0$)
2. $\frac{d}{dx}\left[ {\ln f(x)} \right] = \frac{f'(x)}{f(x)}$ (for $f(x) > 0$)
3. $\frac{d}{dx}\left[ {a^x} \right] = \frac{d}{dx}\left[ e^{x\ln a} \right] = e^{x\ln a} \cdot \ln a = (\ln a){a^x}$
4. $\frac{d}{dx}\left[ {{a^{f(x)}}} \right] = \frac{d}{{dx}}\left[ {{e^{f(x)\ln a}}} \right] = {e^{f(x)\ln a}} \cdot \ln a \cdot f'(x) = (\ln a){a^{f(x)}} \cdot f'(x)$
5. $\frac{d}{dx}\left[ {{{\log }_a}x} \right] = \frac{d}{dx}\left[ {\frac{\ln x}{\ln a}} \right] = \frac{1}{(\ln a)x}$
6. $\frac{d}{{dx}}\left[ {{\log }_a f(x)} \right] = \frac{d}{dx}\left[ {\frac{\ln f(x)}{\ln a}} \right] = \frac{f'(x)}{(\ln a)f(x)}$
Example 1 (formula 3):
$\frac{d}{dx}\left[ {2^x} \right] = \frac{d}{dx}\left[ {e^{x\ln 2}} \right] = {e^{x\ln 2}} \cdot \ln 2 = (\ln 2){2^x}$
Example 2 (formula 4):
$\frac{d}{dx}\left[ {{5^{3{x^2} + 1}}} \right] = \frac{d}{dx}\left[ {{e^{(3{x^2} + 1)\ln 5}}} \right] = {e^{(3{x^2} + 1)\ln 5}} \cdot (6x)(\ln 5) = 6x(\ln 5){5^{3{x^2} + 1}}$
Example 3 (formula 5):
$\frac{d}{dx}\left[ {{{\log }_5}x} \right] = \frac{d}{dx}\left[ {\frac{\ln x}{\ln 5}} \right] = \frac{1}{(\ln 5)x}$
Example 4 (formula 6):
$\frac{d}{dx}\left[ {{{\log }_2}({x^3} + 1)} \right] = \frac{d}{{dx}}\left[ {\frac{{\ln ({x^3} + 1)}}{{\ln 2}}} \right] = \frac{{3{x^2}}}{{(\ln 2)({x^3} + 1)}}$
Use logarithmic differentiation to differentiate the function $f(x) = x^x$.
$$ \begin{aligned} f(x) &= {x^x} \Rightarrow \color{blue}{\ln f(x) = \ln {x^x} = x\ln x}\\ \color{blue}{(\ln f(x))'} &= \frac{f'(x)}{f(x)} = x \cdot \frac{1}{x} + 1 \cdot \ln x = 1 + \ln x \\ &\Rightarrow \ln f'(x) = f(x) (1 + \ln x ) = x^x (1 + \ln x) \\ &\Rightarrow \ln f(x) = \ln {x^x} = x\ln x \\ \frac{f'(x)}{f(x)} &= x \cdot \frac{1}{x} + 1 \cdot \ln x = 1 + \ln x \\ &\Rightarrow f'(x) = f(x) (1 + \ln x) = x^x (1 + \ln x) \end{aligned} $$