use either product, quotient or chain rule to calcuate d/dx squareroot of 2x+ (3x +4x^2)^3 

Given the function:

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

We need to find d/dx or
f'(x).

We will use the chair rule to find the
derivative.

f'(x) = [sqrt(2x)]' +
[(3x+4x^2)^3]'

We know that sqrt2x= (2x)^1/2 = sqrt2 *
(x^1/2)

==> (2x^1/2)' = sqrt2 * (1/2)*x^-1/2 =
sqrt2/ 2sqrtx

Now we will differentiate between
brackets.

==> Let u=
3x+4x^2

==> du = 3+ 8x
d

==> f'(x) = sqrt2/2sqrtx +
(u^3)'

                = sqrt2/2sqrtx+ 3u^2
du

Now we will
substitute:

==> f'(x) = sqrt2/2sqrtx + 3(3x+4x^2)^2
* (3+8x)

               = 1/sqrt2x +
(9+24x)(3x+4x^2)^2

==> d/dx = 1/sqrt2x
+ (9+24x)(3x+4x^2)^2