f(x) = (x^2 + 3x -2)^2   find f'(1)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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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

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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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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