How do you find the derivative of squareroot[ (x + 1)^2 + (4 +4/x)^2 ]?

Question

Whatever the derivative is i need to set it equal to zero to optimize it for the minimum.

Answer 1

write the equation as: -

((x+1)^2+(4+4.x^-1)^2)^0.5

this may be done by the function of a function rule: -
dy/dx = (du/dx).(dy.du)

where the equation as: -
(u)^0.5

thus
u = (x+1)^2 + (4+4.x^-1)^2

Hence dy/du= 0.5u^-0.5

And du/dx may be reduced by the same function of a function method:-

Thus: -

du/dx = 2.(x+1) -(4.x^-2).2.(4+4/x)

Hence the overall derivative is: -

0.5.(2.(x+1)-(8.x-^2).(4+4/x))
_____________________
(( (x+1)^2 + (4+4.x^-1)^2)^0.5)

Answer 2

2(x+1) + 2(4+4/x)(-4/x^2)
=2x +2 -8/x^2 -16/x^3

Answer 3

use http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=mathcom&s1=calculus&s2=differentiate&s3=basic