What is the derivative of $y=(1+x)^(1/x)$ ?

Question

7678 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T14:16:00

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

$=[(1+x)^(1/x-1){x-(1+x)ln(1+x)}]/x^2$

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

$:. lny=ln(1+x)^(1/x)$

$:. lny=1/xln(1+x)=(ln(1+x))/x$

$:. d/dxlny=d/dx{(ln(1+x))/x}$

$:. d/dylny*dy/dx={xd/dxln(1+x)-ln(1+x)d/dxx}/x^2$

Here, we have used the Chain Rule & the Quotient Rule.

$:. 1/ydy/dx={x/(1+x)-ln(1+x)}/x^2={x-(1+x)ln(1+x)}/(x^2(1+x)$

$:. dy/dx=[y{x-(1+x)ln(1+x)}]/(x^2(1+x)$

Using $(star)$ , we get,

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

$=[(1+x)^(1/x-1){x-(1+x)ln(1+x)}]/x^2$

What is the derivative of y=(1+x)^(1/x)y=(1+x)^(1/x)?