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

2 Answers
Jan 31, 2016

(x+1)^x(x/(x+1)+lnx)(x+1)x(xx+1+lnx)

Explanation:

y=(x+1)^xy=(x+1)x

lets use lnln in the two sides of the equation,

lny=xln(x+1)lny=xln(x+1)

now lets differentiate the two sides in respect of xx

1/y(dy)/(dx)=xd/(dx)(ln(x+1))+lnxd/(dx)(x)1ydydx=xddx(ln(x+1))+lnxddx(x)

or,1/yd/dx=x*1/(x+1)d/dx(x+1)+lnxor,1yddx=x1x+1ddx(x+1)+lnx

or,1/y(dy)/(dx)=x/(x+1)+lnxor,1ydydx=xx+1+lnx

or,(dy)/(dx)=y(x/(x+1)+lnx)or,dydx=y(xx+1+lnx)

or,(dy)/(dx)=(x+1)^x(x/(x+1)+lnx)or,dydx=(x+1)x(xx+1+lnx)

Jan 31, 2016

(x+1)^x(ln(x+1)+x/(x+1))(x+1)x(ln(x+1)+xx+1)

Explanation:

I think the easiest way to do these kinds of problems (with a variable power) without memorizing the formula is through implicit differentiation.

y=(x+1)^xy=(x+1)x

Take the natural logarithm of both sides.

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

Rewrite using logarithm rules.

ln(y)=xln(x+1)ln(y)=xln(x+1)

Differentiate both sides with respect to xx.

The left side will spit out a dy/dxdydx term, thanks to the chain rule. The right side will require the product rule.

1/y(dy/dx)=ln(x+1)d/dx(x)+xd/dx(ln(x+1))1y(dydx)=ln(x+1)ddx(x)+xddx(ln(x+1))

Note that differentiating ln(x+1)ln(x+1) requires chain rule as well, but since d/dx(x+1)=1ddx(x+1)=1 it doesn't make a visible difference.

1/y(dy/dx)=ln(x+1)+x(1/(x+1))d/dx(x+1)1y(dydx)=ln(x+1)+x(1x+1)ddx(x+1)

1/y(dy/dx)=ln(x+1)+x/(x+1)1y(dydx)=ln(x+1)+xx+1

Now, solve for dy/dxdydx by multiplying both sides by yy. However, remember that y=(x+1)^xy=(x+1)x.

dy/dx=(x+1)^x(ln(x+1)+x/(x+1))dydx=(x+1)x(ln(x+1)+xx+1)