Exponential နဲ႔ Logarithmic Function ေတြကို differentiate လုပ္ရင္ ေအာက္ပါ Formula ေတြ သိထားဖို႔ လိုပါတယ္။
| $ \displaystyle \begin{array}{*{20}{l}} \begin{array}{l}(1)\ \frac{d}{{dx}}({{a}^{x}})={{a}^{x}}\cdot \ln a,\ \\\ \ \ \ \ \operatorname{where}\ a>0\ \operatorname{and}\ a\ne 1.\end{array} \\ {} \\ {(2)\ \frac{d}{{dx}}({{e}^{x}})={{e}^{x}}} \\ {} \\ {(3)\ \frac{d}{{dx}}({{{\log }}_{b}}x)=\frac{1}{x}{{{\log }}_{b}}e} \\ {} \\ {(4)\ \frac{d}{{dx}}(\ln x)=\frac{1}{x}} \end{array}$ |
ဒါ့အျပင္ logarithm ရဲ့ basic rule တစ္ခ်ိဳ႕ျဖစ္တယ္ ေအာက္ပါ ဥပေဒသေတြကိုလည္း သိထားရပါမယ္။
| $ \displaystyle \begin{array}{*{20}{l}} {(1)\ \ {{{\log }}_{b}}{{b}^{x}}=x} \\ {} \\ {(2)\ \ {{{\log }}_{b}}{{a}^{x}}=x{{{\log }}_{b}}a} \\ {} \\ \begin{array}{l}(3)\ \ {{\log }_{b}}(xy)={{\log }_{b}}x+{{\log }_{b}}y\\\\(4)\ \ {{\log }_{b}}(\frac{x}{y})={{\log }_{b}}x-{{\log }_{b}}y\end{array} \end{array}$ |
Question : If $ \displaystyle y={{x}^{x}}$, prove that $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{y}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}-\frac{y}{x}=0$.
Solution
$ \displaystyle \begin{array}{l}\ \ \ y={{x}^{x}}\\\\\therefore \ln y=\ln {{x}^{x}}\\\\\therefore \ln y=x\ln x\end{array}$
$ \displaystyle \operatorname{Differentiate \ with \ respect \ to \ x}$
$ \displaystyle \ \ \frac{1}{y}\ \frac{{dy}}{{dx}}=x\frac{d}{{dx}}(\ln x)+\ln x\frac{d}{{dx}}(x)$
$ \displaystyle \ \ \frac{1}{y}\ \frac{{dy}}{{dx}}=x\cdot \frac{1}{x}+\ln x(1)$
$ \displaystyle \frac{1}{y}\ \frac{{dy}}{{dx}}=1+\ln x$
$ \displaystyle \operatorname{Differentiate \ again \ with \ respect \ to \ x}$
$ \displaystyle \ \ \frac{1}{y}\cdot \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)+\frac{{dy}}{{dx}}\cdot \frac{d}{{dx}}\left( {\frac{1}{y}} \right)=\frac{d}{{dx}}(1)+\frac{d}{{dx}}\left( {\ln x} \right)$
$ \displaystyle \ \ \frac{1}{y}\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{{{{y}^{2}}}}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=0+\frac{1}{x}$
$ \displaystyle \ \ \frac{1}{y}\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{{{{y}^{2}}}}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=\frac{1}{x}$
$ \displaystyle \operatorname{Multiplying \ both \ sides \ with \ y,}$
$ \displaystyle \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{y}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=\frac{y}{x}$
$ \displaystyle \therefore \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{y}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}-\frac{y}{x}=0$
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