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

The derivative of \sqrt[5]{x}+2sqrt(x^5) is (1+25x^{23/10})/(5x^{4/5)}

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

The first derivative of \sqrt[5]{x}+2sqrt(x^5) is (1+25x^{23/10})/(5x^{4/5)}

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\frac{d}{d x} (5 x + 2 x^{5})

\frac{1 + 25 x ^{\frac{23}{10}}}{5 x ^{\frac{4}{5}}}

Solution steps

\frac{d}{d x} (5 x + 2 x^{5})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (5 x) + \frac{d}{d x} (2 x^{5})

\frac{d}{d x} (5 x) = \frac{1}{5 x ^{\frac{4}{5}}}

\frac{d}{d x} (2 x^{5}) = 5 x^{\frac{3}{2}}

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

Simplify

\frac{1}{5 x ^{\frac{4}{5}}} + 5 x^{\frac{3}{2}} : \frac{1 + 25 x ^{\frac{23}{10}}}{5 x ^{\frac{4}{5}}}

= \frac{1 + 25 x ^{\frac{23}{10}}}{5 x ^{\frac{4}{5}}}

f (x) = (5 x^{3} + 6 x^{2} + 4)^{8}

y = sin (sin (x)), (2 π, 0)

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 6 e^{- x}

\int x sin (\frac{1}{10} x) d x

\int x + 1 ln (x + 1) d x

What is the derivative of \sqrt[5]{x}+2sqrt(x^5) ?
The derivative of \sqrt[5]{x}+2sqrt(x^5) is (1+25x^{23/10})/(5x^{4/5)}
What is the first derivative of \sqrt[5]{x}+2sqrt(x^5) ?
The first derivative of \sqrt[5]{x}+2sqrt(x^5) is (1+25x^{23/10})/(5x^{4/5)}