What is the derivative of y=sqrt(x)+\sqrt[7]{x} ?

The derivative of y=sqrt(x)+\sqrt[7]{x} is 1/(2sqrt(x))+1/(7x^{6/7)}

What is the first derivative of y=sqrt(x)+\sqrt[7]{x} ?

The first derivative of y=sqrt(x)+\sqrt[7]{x} is 1/(2sqrt(x))+1/(7x^{6/7)}

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derivative of

\frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

= \frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}}

10 t^{3} e^{- 2 t}

y = \frac{1}{2} x^{3} + 2, y = x + 1, 0, 2

f (x) = x^{2} + x, (- 2, 2)

\int_{3}^{7} \frac{1}{6 x + 7} d x

\frac{\partial}{\partial u} (\frac{v}{u - v})

What is the derivative of y=sqrt(x)+\sqrt[7]{x} ?
The derivative of y=sqrt(x)+\sqrt[7]{x} is 1/(2sqrt(x))+1/(7x^{6/7)}
What is the first derivative of y=sqrt(x)+\sqrt[7]{x} ?
The first derivative of y=sqrt(x)+\sqrt[7]{x} is 1/(2sqrt(x))+1/(7x^{6/7)}