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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int t^{5} d t

\frac{1}{s ( s + 4 ) ^{2}}

\frac{d}{d x} ((3 x^{2} + 5)^{3} (3 x - 1)^{2})

\int (1 + x^{2})^{- 1} d x

f (x) = (e^{5 x} - 4)^{2}, (0, 9)

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