What is the d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) ?

The d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) is 1/(3t^{2/3)}+9sqrt(t)

What is the first d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) ?

The first d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) is 1/(3t^{2/3)}+9sqrt(t)

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Popular >

\frac{1}{3 t ^{\frac{2}{3}}} + 9 t

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d t} (6 t^{3}) = 9 t

= \frac{1}{3 t ^{\frac{2}{3}}} + 9 t

\frac{x ^{2}}{x - 1}

f (x) = \frac{1}{ln ( x )}

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

\frac{d}{d x} (cos (2 x) sin (3 x))

\int (7 x^{6} - \frac{6}{x ^{6}}) d x

What is the d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) ?
The d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) is 1/(3t^{2/3)}+9sqrt(t)
What is the first d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) ?
The first d/(dt)(\sqrt[3]{t}+6sqrt(t^3)) is 1/(3t^{2/3)}+9sqrt(t)