How do you differentiate $y=(1+x)^(1/x)$ ?

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Answer 1 · 2017-03-31T20:53:26

$dy/dx = (1+x)^(1/x) {1/(x(1+x)) - ln(1+x)/x^2}$

We have:

$y = (1+x)^(1/x)$

First note that the function is not defined when $x=0$ . We can take (natural) logarithms of both sides to get:

$lny = ln{(1+x)^(1/x)}$
$" " = 1/x \ ln(1+x)$
$" " = ln(1+x)/x$

Differentiating implicitly and applying the quotient rule and the chain rule gives:

$1/y dy/dx = { (x)(1/(1+x)) - (ln(1+x))(1) } / (x)^2$
$" " = { x/(1+x) - ln(1+x) } / (x)^2$
$" " = 1/(x(1+x)) - ln(1+x)/x^2$

And so:

$dy/dx = y {1/(x(1+x)) - ln(1+x)/x^2}$
$" " = (1+x)^(1/x) {1/(x(1+x)) - ln(1+x)/x^2}$

Answer 2 · 2017-03-31T20:54:14

The answer is $=(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)$

We diiferentiate using logs.

$y=(1+x)^(1/x)$

$ln(y)=ln((1+x)^(1/x))$

$lny=1/x*ln(1+x)$

Differentiating

$(lny)'=(1/x*ln(1+x))'$

$1/y*dy/dx=1/x*1/(1+x)-1/x^2*ln(1+x)$

$dy/dx=y(1/(x(1+x))-1/x^2ln(1+x))$

$dy/dx=(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)$

How do you differentiate y=(1+x)^(1/x)y=(1+x)^(1/x)?