What is the d/(da)((a^2-2a+a)/9) ?

The d/(da)((a^2-2a+a)/9) is (2a-1)/9

What is the first d/(da)((a^2-2a+a)/9) ?

The first d/(da)((a^2-2a+a)/9) is (2a-1)/9

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\frac{d}{d a} (\frac{a ^{2} - 2 a + a}{9})

\frac{2 a - 1}{9}

Solution steps

\frac{d}{d a} (\frac{a ^{2} - 2 a + a}{9})

Take the constant out:

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

= \frac{1}{9} \frac{d}{d a} (a^{2} - 2 a + a)

Apply the Sum/Difference Rule:

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

= \frac{1}{9} (\frac{d}{d a} (a^{2}) - \frac{d}{d a} (2 a) + \frac{d a}{d a})

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

\frac{d}{d a} (2 a) = 2

\frac{d a}{d a} = 1

= \frac{1}{9} (2 a - 2 + 1)

Simplify

\frac{1}{9} (2 a - 2 + 1) : \frac{2 a - 1}{9}

= \frac{2 a - 1}{9}

What is the d/(da)((a^2-2a+a)/9) ?
The d/(da)((a^2-2a+a)/9) is (2a-1)/9
What is the first d/(da)((a^2-2a+a)/9) ?
The first d/(da)((a^2-2a+a)/9) is (2a-1)/9