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Question Number 24227 by tawa tawa last updated on 14/Nov/17
Commented by tawa tawa last updated on 14/Nov/17
please help.
$$\mathrm{please}\:\mathrm{help}. \\ $$
Answered by $@ty@m last updated on 14/Nov/17
x^y =e^(x−y) taking natural log ln x^y =ln e^(x−y) yln x=x−y differentiating (y/x)+ln x(dy/dx)=1−(dy/dx) (dy/dx)(1+ln x)=1−(y/x) (dy/dx)=((x−y)/(x(1+ln x)))
$${x}^{{y}} ={e}^{{x}−{y}} \\ $$$${taking}\:{natural}\:{log} \\ $$$$\mathrm{ln}\:{x}^{{y}} =\mathrm{ln}\:{e}^{{x}−{y}} \\ $$$${y}\mathrm{ln}\:{x}={x}−{y} \\ $$$${differentiating} \\ $$$$\frac{{y}}{{x}}+\mathrm{ln}\:{x}\frac{{dy}}{{dx}}=\mathrm{1}−\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\left(\mathrm{1}+\mathrm{ln}\:{x}\right)=\mathrm{1}−\frac{{y}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}−{y}}{{x}\left(\mathrm{1}+\mathrm{ln}\:{x}\right)} \\ $$
Commented by tawa tawa last updated on 16/Nov/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tawa tawa last updated on 14/Nov/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by abwayh last updated on 16/Nov/17
can we solve another way x^y =e^(x−y) ⇒e^y x^y =e^x ⇒(x e)^y =e^x (by natural log) ln (x e)^y =ln e^x yln (x e) =x y(ln x+1)=x y=(x/(ln x+1)) ⇒ (dy/dx) =((ln x+1−(x.(1/x)))/((ln x+1)^2 )) (dy/dx)=((ln x)/((ln x+1)^2 ))
$$\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{another}\:\mathrm{way} \\ $$$$\mathrm{x}^{\mathrm{y}} =\mathrm{e}^{\mathrm{x}−\mathrm{y}} \Rightarrow\mathrm{e}^{\mathrm{y}} \:\mathrm{x}^{\mathrm{y}} =\mathrm{e}^{\mathrm{x}} \:\Rightarrow\left(\mathrm{x}\:\mathrm{e}\right)^{\mathrm{y}} =\mathrm{e}^{\mathrm{x}} \\ $$$$\left(\mathrm{by}\:\mathrm{natural}\:\mathrm{log}\right) \\ $$$$\mathrm{ln}\:\left(\mathrm{x}\:\mathrm{e}\right)^{\mathrm{y}} =\mathrm{ln}\:\mathrm{e}^{\mathrm{x}} \: \\ $$$$\mathrm{yln}\:\left(\mathrm{x}\:\mathrm{e}\right)\:=\mathrm{x} \\ $$$$\mathrm{y}\left(\mathrm{ln}\:\mathrm{x}+\mathrm{1}\right)=\mathrm{x} \\ $$$$\mathrm{y}=\frac{\mathrm{x}}{\mathrm{ln}\:\mathrm{x}+\mathrm{1}}\:\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\frac{\mathrm{ln}\:\mathrm{x}+\mathrm{1}−\left(\mathrm{x}.\frac{\mathrm{1}}{\mathrm{x}}\right)}{\left(\mathrm{ln}\:\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ln}\:\mathrm{x}}{\left(\mathrm{ln}\:\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by $@ty@m last updated on 16/Nov/17
Its same
$${Its}\:{same} \\ $$
Answered by $@ty@m last updated on 14/Nov/17
∫x^n ln xdx Integrating by parts =ln x∫x^n dx−∫(x^(n+1) /(x.(n+1)))dx =(x^(n+1) /(n+1))ln x−∫(x^n /(n+1)) =(x^(n+1) /(n+1))ln x−(x^(n+1) /((n+1)^2 ))+C
$$\int{x}^{{n}} \mathrm{ln}\:{xdx} \\ $$$${Integrating}\:{by}\:{parts} \\ $$$$=\mathrm{ln}\:{x}\int{x}^{{n}} {dx}−\int\frac{{x}^{{n}+\mathrm{1}} }{{x}.\left({n}+\mathrm{1}\right)}{dx} \\ $$$$=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mathrm{ln}\:{x}−\int\frac{{x}^{{n}} }{{n}+\mathrm{1}} \\ $$$$=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mathrm{ln}\:{x}−\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+{C} \\ $$
Commented by tawa tawa last updated on 14/Nov/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by $@ty@m last updated on 14/Nov/17
lny=lim_(x→0) ((1−cos x)/x^2 ) ln y=lim_(x→0) ((2sin^2 (x/2))/x^2 ) ln y=lim_(x→0) ((2sin^2 (x/2))/(((x/2))^2 ×4)) ln y=(1/2) y=(√e)
$${lny}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:{y}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:{y}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} ×\mathrm{4}} \\ $$$$\mathrm{ln}\:{y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}=\sqrt{{e}} \\ $$
Commented by tawa tawa last updated on 16/Nov/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tawa tawa last updated on 14/Nov/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tawa tawa last updated on 14/Nov/17
It remain the mean value theorem sir.
$$\mathrm{It}\:\mathrm{remain}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{value}\:\mathrm{theorem}\:\mathrm{sir}. \\ $$
Commented by abwayh last updated on 16/Nov/17
can we use L′ Hopital rule;sir ln y=lim_(x→0) ((1−cos x)/x^2 ) ⇒^(L′Hopital) ln y=lim_(x→0) ((sin x)/(2x)) ⇒ ln y=lim_(x→0) ((cos x)/2) ln y=(1/2)
$$\mathrm{can}\:\mathrm{we}\:\mathrm{use}\:\:\:\mathrm{L}'\:\mathrm{Hopital}\:\mathrm{rule};\mathrm{sir} \\ $$$$\mathrm{ln}\:\mathrm{y}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\:\overset{\mathrm{L}'\mathrm{Hopital}} {\Rightarrow}\:\:\:\mathrm{ln}\:\mathrm{y}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2x}}\:\Rightarrow \\ $$$$\mathrm{ln}\:\mathrm{y}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by $@ty@m last updated on 16/Nov/17
yes provided it is taught in the class
$${yes} \\ $$$${provided}\:{it}\:{is}\:{taught}\:{in}\:{the}\:{class} \\ $$

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