f(x) = (x^2 + 3x -2)^2 find f'(1)
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f(x) = (x^2 + 3x -2)^2
Let us calculate f'(x)
f'(x) = 2(x^2 + 3x -2) *(2x + 3)
= (2x^2 + 6x -4 )(2x + 3)
f'(1) = (2 + 6 -4)(2 +3)
= 4*5 = 20
==> f'(1) = 20
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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
To find f'(1). f(x) = (x^2+3x-2)^2.
Solution:
We know that if f(x) = u (v(x) , then
f'(x) = (u(v(x)) ' = u'(v(x))* v'(x).
Here f(x) = (x^2+3x-2)^2
(u(x)+v(x))' = u'(x)+v'(x).
(kx^n)' = knx^(n-1).
Here,
v(x) = x^2+3x-2, u(v) = v^2.
Therefore f'(x) = u'(v)*v'x) = 2(x^2+3x-2)*(x^2+3x-2)'
So,
f'(x) {(x^2+3x-2)^2 }' = 2(x^2+3x-2)^(2-1)*{ (x^2)' +(3x)'-(2)'}
= 2(x^2+3x-2)(2x+3).
f'(1) = 2(1^2+3*1-2)(2*1+3)
f'(1) = 2*2*5 = 20
Given:
f(x) = (x^2 + 3x - 2)^2
Let:
y = (x^2 + 3x - 2)^2
t = x^2 + 3x - 2
Then:
y = t^2
and:
f'(x) = dy/dx = (dy/dt)(dt/dx)
dy/dt = 2t
= 2(x^2 + 3x - 2)
= 2x^2 + 6x - 4
dt/dx = 2x + 3
f'(x) = (dy/dt)(dt/dx)
= (2x^2 + 6x - 4)(2x + 3)
substituting the value of x = 1
f'(1) = 2*1^2 + 6*1 - 4)(2*1 + 3)
= (2 + 6 - 4)(2 + 3)
= 4*5 = 20
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