The correct option is B e e^x^ex^e 1+x^e^xe^x ( 1/x+log x )+e^x^xx^x(1+log x)y=e^x^e+x^e^x+e^x^x y=I+II+III II=e^x logx III=xloge^x dy/dx=e^x^e ex^e 1+e^x ...

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Byju's Answer

Standard XII

Mathematics

Logarithms

d/d xexe+xex+...

Question

$\frac{d}{d x} {e^{x^{e}} + x^{e^{x}} + e^{x^{x}}}$ =

e^{x^{e}} + x^{e^{x}} + e^{x^{x}}

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x^{2} e^{x^{e}} + e^{x} x^{e^{x}} + x^{x} e^{x^{x}}

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e e^{x^{e}} x^{e - 1} + x^{e^{x}} e^{x} (\frac{1}{x} + l o g x) + e^{x^{x}} x^{x} (1 + l o g x)

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Solution

The correct option is C $e e^{x^{e}} x^{e - 1} + x^{e^{x}} e^{x} (\frac{1}{x} + l o g x) + e^{x^{x}} x^{x} (1 + l o g x)$
$y = e^{x^{e}} + x^{e^{x}} + e^{x^{x}} y = I + I I + I I I I I = e^{x} l o g x I I I = x l o g e^{x} \frac{d y}{d x} = e^{x^{e}} e x^{e - 1} + e^{x} (\frac{1}{x}) + l o g x (e^{x}) + x \frac{1}{e^{x}} + l o g e^{x} (1) = e e^{x^{e}} x^{e - 1} + \frac{e^{x}}{x} + e^{x} l o g x + l o g e^{x}$

Suggest Corrections

ddx{exe+xex+exx} =

The correct option is C eexexe−1+xexex(1x+log x)+exxxx(1+log x)y=exe+xex+exxy=I+II+IIIII=exlogx III=xlogexdydx=exeexe−1+ex(1x)+logx(ex)+x1ex+logex(1)=eexexe−1+exx+exlogx+logex

$\frac{d}{d x} {e^{x^{e}} + x^{e^{x}} + e^{x^{x}}}$ =