f(x) = (x^2 + 3x -2)^2 find f'(1)

To calculate the value of the derivative of the function,
for x = 1, we'll have to differentiate the function.

We
notice that we have to calculate the derivative of a composed
function.

Let's suppose that u(v) = v^2 and v(x) = x^2 + 3x
-2

So, [u(v(x))]' =
u'(v)*v'(x)

[u(v(x))]' = (v^2)'*(x^2 + 3x
-2)'

[u(v(x))]' = 2v*(2x+3), where v(x) = x^2 + 3x
-2

[u(v(x))]' = 2(x^2 + 3x
-2)*(2x+3)

f'(1) = [u(v(1)]' = 2*(1^2 + 3*1
-2)*(2*1+3)

f'(1) = [u(v(1)]' =
2*2*5

f'(1) = [u(v(1)]' =
20