What is the derivative of sqrt(x)+\sqrt[5]{x}+1/7 ?

The derivative of sqrt(x)+\sqrt[5]{x}+1/7 is 1/(2sqrt(x))+1/(5x^{4/5)}

What is the first derivative of sqrt(x)+\sqrt[5]{x}+1/7 ?

The first derivative of sqrt(x)+\sqrt[5]{x}+1/7 is 1/(2sqrt(x))+1/(5x^{4/5)}

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

Solution steps

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (\frac{1}{7}) = 0

= \frac{1}{2 x} + \frac{1}{5 x ^{\frac{4}{5}}} + 0

Simplify

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

x \to \infty lim (e)

\int_{0}^{1} (x + 6) (x - 7) d x

\frac{d ^{2}}{d x ^{2}} (\frac{1 - 4 x}{1 + 4 x})

\frac{d y}{d x} = \frac{e ^{x}}{y + 2}

What is the derivative of sqrt(x)+\sqrt[5]{x}+1/7 ?
The derivative of sqrt(x)+\sqrt[5]{x}+1/7 is 1/(2sqrt(x))+1/(5x^{4/5)}
What is the first derivative of sqrt(x)+\sqrt[5]{x}+1/7 ?
The first derivative of sqrt(x)+\sqrt[5]{x}+1/7 is 1/(2sqrt(x))+1/(5x^{4/5)}