What is the derivative of y=4x^3-4x^2+x-1 ?

The derivative of y=4x^3-4x^2+x-1 is 12x^2-8x+1

What is the first derivative of y=4x^3-4x^2+x-1 ?

The first derivative of y=4x^3-4x^2+x-1 is 12x^2-8x+1

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derivative of

y = 4 x^{3} - 4 x^{2} + x - 1

12 x^{2} - 8 x + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (4 x^{3}) = 12 x^{2}

\frac{d}{d x} (4 x^{2}) = 8 x

\frac{d x}{d x} = 1

\frac{d}{d x} (1) = 0

= 12 x^{2} - 8 x + 1 - 0

Simplify

= 12 x^{2} - 8 x + 1

ln (\frac{( x + 1 )}{( x - 1 )})

\frac{d}{d t} (3 t^{2} + 2 t + 1)

\frac{x ^{2}}{16}

\frac{( 1 + x )}{x}

b (1 - sin (x))

What is the derivative of y=4x^3-4x^2+x-1 ?
The derivative of y=4x^3-4x^2+x-1 is 12x^2-8x+1
What is the first derivative of y=4x^3-4x^2+x-1 ?
The first derivative of y=4x^3-4x^2+x-1 is 12x^2-8x+1