What is the derivative of (x^2+3x/7) ?

The derivative of (x^2+3x/7) is 1/7 (2x+3)

What is the first derivative of (x^2+3x/7) ?

The first derivative of (x^2+3x/7) is 1/7 (2x+3)

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\frac{d}{d x} (\frac{x ^{2} + 3 x}{7})

\frac{1}{7} (2 x + 3)

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 3 x}{7})

Take the constant out:

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

= \frac{1}{7} \frac{d}{d x} (x^{2} + 3 x)

Apply the Sum/Difference Rule:

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

= \frac{1}{7} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (3 x))

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

\frac{d}{d x} (3 x) = 3

= \frac{1}{7} (2 x + 3)

\frac{d}{d s} (\frac{s}{s ^{2} + 4})

e^{3 x} (3 cos (4 x) - 4 sin (4 x))

\int \frac{- 5 x ^{2}}{4 x - x ^{2}} d x

f (x) = 8 x^{4} - 3 x^{3} + 4 x

\frac{d}{d x} (cos (sin (x^{4})))

What is the derivative of (x^2+3x/7) ?
The derivative of (x^2+3x/7) is 1/7 (2x+3)
What is the first derivative of (x^2+3x/7) ?
The first derivative of (x^2+3x/7) is 1/7 (2x+3)