What is the derivative of f(x)=((2x^2-3x+1)^2)/4+2 ?

The derivative of f(x)=((2x^2-3x+1)^2)/4+2 is 1/2 (4x-3)(2x^2-3x+1)

What is the first derivative of f(x)=((2x^2-3x+1)^2)/4+2 ?

The first derivative of f(x)=((2x^2-3x+1)^2)/4+2 is 1/2 (4x-3)(2x^2-3x+1)

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derivative of

f (x) = \frac{( 2 x ^{2} - 3 x + 1 ) ^{2}}{4} + 2

\frac{1}{2} (4 x - 3) (2 x^{2} - 3 x + 1)

Solution steps

\frac{d}{d x} (\frac{( 2 x ^{2} - 3 x + 1 ) ^{2}}{4} + 2)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{( 2 x ^{2} - 3 x + 1 ) ^{2}}{4}) + \frac{d}{d x} (2)

\frac{d}{d x} (\frac{( 2 x ^{2} - 3 x + 1 ) ^{2}}{4}) = \frac{1}{2} (2 x^{2} - 3 x + 1) (4 x - 3)

\frac{d}{d x} (2) = 0

= \frac{1}{2} (2 x^{2} - 3 x + 1) (4 x - 3) + 0

Simplify

= \frac{1}{2} (4 x - 3) (2 x^{2} - 3 x + 1)

n = 1 \sum \infty (- x)^{n + 1}

9 ln (5 - 8 x^{2})

\frac{d y}{d t} = 3 + 2 y

y = x^{3} - 3 x - 4

\frac{d y}{d x} = \frac{2 x ( y - 1 )}{x ^{2} + 3}, y (1) = 9

What is the derivative of f(x)=((2x^2-3x+1)^2)/4+2 ?
The derivative of f(x)=((2x^2-3x+1)^2)/4+2 is 1/2 (4x-3)(2x^2-3x+1)
What is the first derivative of f(x)=((2x^2-3x+1)^2)/4+2 ?
The first derivative of f(x)=((2x^2-3x+1)^2)/4+2 is 1/2 (4x-3)(2x^2-3x+1)