What is the derivative of (x^2+40x/5) ?

The derivative of (x^2+40x/5) is 1/5 (2x+40)

What is the first derivative of (x^2+40x/5) ?

The first derivative of (x^2+40x/5) is 1/5 (2x+40)

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\frac{d}{d x} (\frac{x ^{2} + 40 x}{5})

\frac{1}{5} (2 x + 40)

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 40 x}{5})

Take the constant out:

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

= \frac{1}{5} \frac{d}{d x} (x^{2} + 40 x)

Apply the Sum/Difference Rule:

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

= \frac{1}{5} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (40 x))

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

\frac{d}{d x} (40 x) = 40

= \frac{1}{5} (2 x + 40)

What is the derivative of (x^2+40x/5) ?
The derivative of (x^2+40x/5) is 1/5 (2x+40)
What is the first derivative of (x^2+40x/5) ?
The first derivative of (x^2+40x/5) is 1/5 (2x+40)