What is the derivative of \sqrt[5]{x^2}+4sqrt(x^3) ?

The derivative of \sqrt[5]{x^2}+4sqrt(x^3) is 2/(5x^{3/5)}+6sqrt(x)

What is the first derivative of \sqrt[5]{x^2}+4sqrt(x^3) ?

The first derivative of \sqrt[5]{x^2}+4sqrt(x^3) is 2/(5x^{3/5)}+6sqrt(x)

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\frac{2}{5 x ^{\frac{3}{5}}} + 6 x

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (4 x^{3}) = 6 x

= \frac{2}{5 x ^{\frac{3}{5}}} + 6 x

\frac{\partial}{\partial y} (x^{2} sin (2 y))

x \to - 6 lim (x - 6 - x)

cos (5 + x^{2})

\frac{\partial}{\partial y} (- arctan (\frac{x}{y}))

\frac{3 ^{2}}{s ( s + 3 ) ^{2}}

What is the derivative of \sqrt[5]{x^2}+4sqrt(x^3) ?
The derivative of \sqrt[5]{x^2}+4sqrt(x^3) is 2/(5x^{3/5)}+6sqrt(x)
What is the first derivative of \sqrt[5]{x^2}+4sqrt(x^3) ?
The first derivative of \sqrt[5]{x^2}+4sqrt(x^3) is 2/(5x^{3/5)}+6sqrt(x)