What is the derivative of 3x^2+sqrt(x)+4 ?

The derivative of 3x^2+sqrt(x)+4 is 6x+1/(2sqrt(x))

What is the first derivative of 3x^2+sqrt(x)+4 ?

The first derivative of 3x^2+sqrt(x)+4 is 6x+1/(2sqrt(x))

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

6 x + \frac{1}{2 x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (x) = \frac{1}{2 x}

\frac{d}{d x} (4) = 0

= 6 x + \frac{1}{2 x} + 0

Simplify

= 6 x + \frac{1}{2 x}

What is the derivative of 3x^2+sqrt(x)+4 ?
The derivative of 3x^2+sqrt(x)+4 is 6x+1/(2sqrt(x))
What is the first derivative of 3x^2+sqrt(x)+4 ?
The first derivative of 3x^2+sqrt(x)+4 is 6x+1/(2sqrt(x))