What is the d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) ?

The d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) is (1+50t^{23/10})/(5t^{4/5)}

What is the first d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) ?

The first d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) is (1+50t^{23/10})/(5t^{4/5)}

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\frac{1 + 50 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d t} (4 t^{5}) = 10 t^{\frac{3}{2}}

= \frac{1}{5 t ^{\frac{4}{5}}} + 10 t^{\frac{3}{2}}

Simplify

\frac{1}{5 t ^{\frac{4}{5}}} + 10 t^{\frac{3}{2}} : \frac{1 + 50 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

= \frac{1 + 50 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

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

y^{^{'''}} - 2 y^{^{'}} + y = 0

\int - 9 sin (3 x) d x

\int 12 x ln (3 x) d x

\frac{d y}{d t} = t y^{\frac{1}{3}}

What is the d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) ?
The d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) is (1+50t^{23/10})/(5t^{4/5)}
What is the first d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) ?
The first d/(dt)(\sqrt[5]{t}+4sqrt(t^5)) is (1+50t^{23/10})/(5t^{4/5)}