Which is the result of differentiating the function f = ( x^2 + 2x + 2 )^2  ?

Expert Answers
hala718 eNotes educator| Certified Educator

f = ( x^2 + 2x + 2 )^2 

assume u = x^2 +2x +2    then,  u'= 2x+2

assume v= x^2       then, v' = 2x

now f = v(u(x))

then f' = (vou)' = [v'(u)](u')

                   = v'(x^2+2x+2) (2x+2)

                   = 2(x^2+2x+2 )(2x+2)

                 

neela | Student

f(x) = (x^2+2x+2)^2.

To find the diferential coeefficient of f(x).

Solution:

We know that d/dx{u(v(x)} = (du/dv) (dv/dx)

Let v(x) = x^2+2x+2. and u(v) = v^2.Then

f'(x) = 2v*dv/dx = 2v *(x^2+2x+2)' = 2v (2x+2) = 4(x^2+2x+2)(x+1), as (d/dx)(nx^n) = k*n*x^(n-1)

 

giorgiana1976 | Student

To calculate the result of the derivative we'll use the chain rule:

We'll note  x^2 + 2x + 2 = u

f(x) = y = (  x^2 + 2x + 2 )^2 = u^2

dy/du=2u

du/dx= (  x^2 + 2x + 2 )' = 2x + 2

dy/dx=(dy/du)*(du/dx)=2*( x^2 + 2x + 2)*(2x + 2)

Another way to solve this problem is to expand the square first and, after expanding, to differentiate the function, with respect to x.

( x^2 + 2x + 2 )^2=x^4+4x^2+4+2(2x^3+2x^2+4x)

[x^4+4x^2+4+2(2x^3+2x^2+4x)]'=4x^3+8x+12x^2+8x+8

[( x^2 + 2x + 2 )^2]'=4x^3+4x^3+16x+8