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

The derivative of y=(x^2-x+1)/(a^3-3) is (2x-1)/(a^3-3)

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

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

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derivative of

y = \frac{x ^{2} - x + 1}{a ^{3} - 3}

\frac{2 x - 1}{a ^{3} - 3}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

\frac{1}{a ^{3} - 3} (2 x - 1 + 0) : \frac{2 x - 1}{a ^{3} - 3}

= \frac{2 x - 1}{a ^{3} - 3}

y = t^{2}

\frac{sec ( x )}{x ^{2}}

\frac{d}{d t} (6 + 3 t^{2} + 2 t)

\frac{d}{d x} (k \cdot x^{2} - 2 x + 1)

sin^{2} (x + 4)

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