IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

c_001

Ques. if x^e^{^y+e^y.......infinity}

then find dy/dx

Note: e^y+e^y.... means e^{y+e^y......infinity} $\;\;\;$

spideyunlimited

Is my answer correct? took log twice and used clever substitution too.

$\frac{dy}{dx} = \frac{log(\frac{x^{1/y}}{y^{1/x}} )(log_y x - 1)}{(logx)^{2}}$

c_001

I am sorry my friend , but the answer is (1-x)/x

you can find this question in Objective Approach -Mathematics by Arihant Publication , Work Book Exercise 7.2

nunknown91

do u mean the ques is like this

${x}^{{e}^{y+{e}^{y+{e}^{y+{e}^{y+\ldots \infty}}}}}$

then $\frac{dy}{dx} = ?$

kkhushh

Hey ! i think the answer given by SPIDEY is all right

even i m getting it likewise .U can check it urself by

1.taking log of given expression

2.u get e^y+e^y.............infinity=lny/lnx

3.then replace exponent of L.H.S. by e^y+lny/lnx

4.differentiate the expression u ll get in 4 n rearrange

i haven't exactly rearranged it cause it seems quite taxing to do so but it more or less assmes what spidey's saying

magiclko

even i am getting somewhat same answer...

c_001

Even i am expecting the same answer but the book says something else . I know the book answer can be wrong but its probability is quite low . There should be some other logic , some other way to do this question .

allamraju

I think the question shud be If x= ${e}^{y+e^y+...}$ ,then find dy/dx

In that case,U get the answer u posted b'coz

lnx=y+x $\Rightarrow$ 1/x=y'+1 $\Rightarrow$ y'= $\frac{1-x}{x}$

atul_sinha89

HEY ALLAM CAN PLS DESCRIBE ABOUT IT ...MORE..

allamraju

Generally,when something of this form ${e}^{y+e^y+...}$ occurs,we take it as ${e}^{y+x}$ since x= ${e}^{y+e^y+...}$ since infinity+1 and infinity are almost same.Thus,

x=e^y+xmeans lnx=y+x

@sinha,Hope u understood.

c_001

Thanks , Allam, you have solved my problem .

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I think the question shud be If x=,then find dy/dx

In that case,U get the answer u posted b'coz

lnx=y+x1/x=y'+1y'=

Generally,when something of this form occurs,we take it as since x=since infinity+1 and infinity are almost same.Thus,

x=ey+x means lnx=y+x

@sinha,Hope u understood.

I think the question shud be If x= ${e}^{y+e^y+...}$ ,then find dy/dx

lnx=y+x $\Rightarrow$ 1/x=y'+1 $\Rightarrow$ y'= $\frac{1-x}{x}$

Generally,when something of this form ${e}^{y+e^y+...}$ occurs,we take it as ${e}^{y+x}$ since x= ${e}^{y+e^y+...}$ since infinity+1 and infinity are almost same.Thus,

x=e^y+xmeans lnx=y+x