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

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Student Comments

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**