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

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

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

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

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

\frac{3 x}{2} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

= \frac{3 x}{2} + 1

n = 1 \sum \infty \frac{1}{n ^{2} + 5}

\int \frac{1}{( 4 + x ^{2} )} d x

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

y = 5 x tan (x), at x = π

f (x) = 3^{x^{5}}

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