What is the derivative of f(x)=\sqrt[3]{x}+sqrt(x) ?

The derivative of f(x)=\sqrt[3]{x}+sqrt(x) is 1/(3x^{2/3)}+1/(2sqrt(x))

What is the first derivative of f(x)=\sqrt[3]{x}+sqrt(x) ?

The first derivative of f(x)=\sqrt[3]{x}+sqrt(x) is 1/(3x^{2/3)}+1/(2sqrt(x))

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derivative of

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

Solution steps

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (x^{2} - 6)

\int_{0}^{2} \frac{2 x}{( x ^{2} + 3 ) ^{2}} d x

(14, 448), (1, y)

\int 4

x \to 0 + lim (\frac{1}{x} - \frac{1}{∣ x ∣})

What is the derivative of f(x)=\sqrt[3]{x}+sqrt(x) ?
The derivative of f(x)=\sqrt[3]{x}+sqrt(x) is 1/(3x^{2/3)}+1/(2sqrt(x))
What is the first derivative of f(x)=\sqrt[3]{x}+sqrt(x) ?
The first derivative of f(x)=\sqrt[3]{x}+sqrt(x) is 1/(3x^{2/3)}+1/(2sqrt(x))