What is the derivative of y=sqrt(x)+\sqrt[6]{x} ?

The derivative of y=sqrt(x)+\sqrt[6]{x} is 1/(2sqrt(x))+1/(6x^{5/6)}

What is the first derivative of y=sqrt(x)+\sqrt[6]{x} ?

The first derivative of y=sqrt(x)+\sqrt[6]{x} is 1/(2sqrt(x))+1/(6x^{5/6)}

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derivative of

\frac{1}{2 x} + \frac{1}{6 x ^{\frac{5}{6}}}

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}{6 x ^{\frac{5}{6}}}

\int \frac{csc ^{4} ( 2 x )}{cot ^{2} ( 2 x )} d x

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

x y^{^{'}} + y = - 9 x y^{2}, y (1) = - 2

x \to x + 2 lim (2 x^{2} + 3 x - 2)

(\frac{sin ( x )}{x})^{^{'}}

What is the derivative of y=sqrt(x)+\sqrt[6]{x} ?
The derivative of y=sqrt(x)+\sqrt[6]{x} is 1/(2sqrt(x))+1/(6x^{5/6)}
What is the first derivative of y=sqrt(x)+\sqrt[6]{x} ?
The first derivative of y=sqrt(x)+\sqrt[6]{x} is 1/(2sqrt(x))+1/(6x^{5/6)}