Calculus : Gradient of Tangent

ေပးထားတဲ့ curve (အစိမ္းေရာင္) က $ \displaystyle y=f(x)$ ျဖစ္ပါတယ္။ ခရမ္းေရာင္ $ \displaystyle PQ$ မ်ဥ္းကေတာ့ $ \displaystyle \text{curve}$ ေပၚမွာရွိတဲ့ အမွတ္ႏွစ္ခုကို ျဖတ္သြားလို႔ $ \displaystyle \text{secant}$ လို႔ ေခၚမယ္။ $ \displaystyle P$ အမွတ္မွာ $ \displaystyle \text{curve}$ ကို ထိသြားတဲ့ အနက္ေရာင္မ်ဥ္းကိုေတာ့ $ \displaystyle \text{tangent}$ လို႔ေခၚမယ္။


Given (ေပးခ်က္) : Curve : $ \displaystyle y=f(x)$


Claim (ရွာရန္) : Gradient (Slope) of tangent at P


Explanation (ရွင္းလင္းခ်က္)


ေပးထားတာ curve : $ \displaystyle y=f(x)$  ရွိတာေၾကာင့္ curve equation ထဲကို $ \displaystyle x$ ေတြထည့္ရင္ $ \displaystyle y$ ရမွာေပါ့။ တနည္းေျပာရင္ curve ေပၚမွာရွိတဲ့ အမွတ္ေတြကို ႀကိဳက္သေလာက္ ရွာႏိုင္ပါတယ္။ curve ေပၚမွာရွိတဲ့ အမွတ္ႏွစ္ကို ျဖတ္ဆြဲရင္ secant ေပါ့။ အမွတ္ႏွစ္မွတ္ သိမွေတာ့ gradient (slope) ကိုလည္း ရွာလို႔ရၿပီေပါ့။


$ \displaystyle \begin{array}{l}\ \ \ y=f(x)\\\\\ \ \ {{y}_{1}}=f({{x}_{1}})\Rightarrow y+\delta y=f(x+\delta x)\\\\\therefore \ \text{Gradient of secant}\ PQ\\\\=\frac{{{{y}_{1}}-y}}{{{{x}_{1}}-x}}\\\\=\frac{{y+\delta y-y}}{{x+\delta x-x}}\ \text{or }\frac{{f(x+\delta x)-f(x)}}{{x+\delta x-x}}\\\\=\frac{{\delta y}}{{\delta x}}\ \text{or }\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array}$


ဒါေပမယ့္ အခုလိုခ်င္တာက gradient of tangent ျဖစ္ပါတယ္။ secant မဟုတ္ဘူး။ ျပသနာက tangent က curve ေပၚမွာ အမွတ္တစ္မွတ္ကိုပဲ ထိသြားတာ၊ ႏွစ္မွတ္မရွိဘူး။ တစ္မွတ္ထဲကို ႏွစ္ကိုယ္ခြဲတြက္ေပါ့ လို႔ေျပာလိုက ေျပာႏိုင္ပါေသးတယ္။ ႏွစ္ကိုယ္ခြဲလိုက္မွေတာ့ $ \displaystyle \frac{{{{y}_{1}}-y}}{{{{x}_{1}}-x}}=\frac{{y-y}}{{{{y}_{1}}-x}}=\frac{0}{0}$ ဆိုတာ indeterminate form ျဖစ္သြားၿပီ ဘာမွ ဆက္လုပ္လို႔မရေတာ့ဘူး။


ေရ တစ္စည္ထဲကို ေရတစ္ခြက္ ေပါင္းထည့္လို႔၊ ေရတစ္စည္ထဲက ေရတစ္ခြက္ ခပ္ထုတ္လိုက္လို႔ ေရ တစ္စည္ကို တိုးလားတယ္ ေလ်ာ့သြားတယ္လို႔ ေျပာေလ့မရွိၾကပါဘူး။ ဘာေၾကာင့္လဲ ဆိုေတာ့ $\displaystyle \frac{{\operatorname{ေရတစ္ခြက္}}}{{\operatorname{ေရတစ္စည္}}}\approx0$ ျဖစ္တာေၾကာင့္ပါ။ သခၤ်ာ႐ွဳေထာင့္က ၾကည့္ရင္ေတာ့ တစ္ခြက္တိုးတိုး တစ္စက္ တိုးတိုး အတိုး ရွိတာေပါ့့။


အလားတူပါပဲ တစ္မွတ္ထဲပဲ ရွိတဲ့ tangent ရဲ့ gradient ကို မရွာႏိုင္ေပမယ့္ အမွတ္ $ \displaystyle P$ နားကို အလြန္နီးကပ္ေနတဲ့ အမွတ္တစ္ခုကို ယူလိုက္ရင္ေတာ့ အမွတ္ႏွစ္ခု ျဖစ္သြားလို႔ gradient ရွာႏိုင္ၿပီေပါ့။  tangent ေတာ့မဟုတ္ဘူး tangent နား အလြန္ကပ္ေနတဲ့ secant ရဲ့ gradient ေပါ့။ Calculus မွာေတာ့ $ \displaystyle Q$ က $ \displaystyle P$ အနားကို လံုေလာက္ေအာင္ နီးကပ္သြားရင္ Gradient of tangent = Gradient of Secant လို႔ သတ္မွတ္ပါတယ္။


ပံုမွာ ျမင္ေတြ႔ရတဲ့ အတိုင္းေပါ့။ $ \displaystyle Q$ က Curve တေလွ်ာက္ $ \displaystyle P$ အနားကို ကပ္သြားဖို႔ $ \displaystyle {{{x}_{1}}}$ ရဲ့ တန္ဖိုး ေလ်ာ႔သြားဖို႔လိုပါတယ္။ $ \displaystyle {{x}_{1}}=x+\delta x$ ျဖစ္တာေၾကာင့္ $ \displaystyle {{{x}_{1}}}$ ရဲ့ တန္ဖိုး ေလ်ာ့သြားဖို႔ ဆိုတာက $ \displaystyle {\delta x}$ တန္ဖိုး ေလ်ာ့သြားမွ ျဖစ္မွာေပါ့။ ပံုမွာ $ \displaystyle {\delta x}$ တန္ဖိုးသတ္မွတ္ထားတဲ့ slider ကို ဘယ္ဘက္ကို ေရႊ႕ၾကည့္ပါ။


$ \displaystyle \begin{array}{*{20}{l}} {\text{When }\delta x\to 0,\ } \\ {} \\ {\text{Gradient of secant}\to \text{Gradient of tangent}} \\ {} \\ {\text{Therefore the gradient of secant approaches }} \\ {\text{the gradient of tangent when }\delta x\ \text{approaches 0}\text{.}} \\ {} \\ \begin{array}{l}\text{By limit notation,}\\\text{ }\end{array} \\ \begin{array}{l}\text{Gradient of tangent =}\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}\\\\\text{Gradient of tangent =}\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array} \end{array}$


Gradient of tangent ကိုေတာ့ သေကၤတ $ \displaystyle \frac{{dy}}{{dx}}$ (သို႔) $ \displaystyle y'$ (သို႔) $ \displaystyle f'(x)$ (သို႔) $\displaystyle \frac{d}{{dx}}\left[ {f(x)} \right]$ ျဖင့္သတ္မွတ္ပါတယ္။ ဒါ့ေၾကာင့္ ...


$ \displaystyle \begin{array}{l}\frac{{dy}}{{dx}}={y}'=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}\\{f}'(x)=\frac{d}{{dx}}\left[ {f(x)} \right]=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array}$

Differentiation from the First Principle

$ \displaystyle \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}$ or $ \displaystyle {f}'(x)=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}$  ဘယ္လို ျဖစ္သြားလဲ...

Problem Study : Deravitive of Exponential and Logarithmic Function

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$ 

Differentiation from the first principles.

Differentiate $ \displaystyle y=\sqrt[3]{x}$ from the first principles.

Solution 

      $ \displaystyle \ \ \ y=\sqrt[3]{x}={{x}^{{\frac{1}{3}}}}$

     $ \displaystyle \therefore y+\delta y={{\left( {x+\delta x} \right)}^{{\frac{1}{3}}}}$ 
      
      $ \displaystyle \therefore \delta y={{\left( {x+\delta x} \right)}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}$

      $ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}}}{{\delta x}}$

      $ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}}}{{\delta x}}\times \frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$  (***)

      $ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{x+\delta x-x}}{{\delta x\left( {{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}} \right)}}$ 

      $ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{1}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$ 

      $ \displaystyle \therefore \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}$

      $ \displaystyle \ \ \ \ \ \ \ \,=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{1}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$ 

     $ \displaystyle \ \ \ \ \ \ \ \,=\frac{1}{{{{x}^{{\frac{2}{3}}}}+{{x}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$

     $ \displaystyle \ \ \ \ \ \ \ \,=\frac{1}{{3{{x}^{{\frac{2}{3}}}}}}$

     Solution ရဲ့ line 5 (***) မွာ indeterminate form ကို ေျပာင္းဖို႔ conjugate နဲ႔ ေျမႇာက္တာ ျဖစ္ပါတယ္။ 

$ \displaystyle {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$  

    ဆိုတဲ့ ပံုေသနည္းကို အသံုးျပဳတာ ျဖစ္ပါတယ္။