What is the derivative of f(x)=sqrt(x)+\sqrt[4]{x} ?

The derivative of f(x)=sqrt(x)+\sqrt[4]{x} is 1/(2sqrt(x))+1/(4x^{3/4)}

What is the first derivative of f(x)=sqrt(x)+\sqrt[4]{x} ?

The first derivative of f(x)=sqrt(x)+\sqrt[4]{x} is 1/(2sqrt(x))+1/(4x^{3/4)}

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derivative of

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

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}{4 x ^{\frac{3}{4}}}

6 \frac{d y}{d x} - 2 y = x y^{4}, y (0) = - 2

(\frac{2 y}{x} + 2 x) + (2 ln (x) - 3) (\frac{d y}{d x}) = 0

\frac{1}{4 ^{x}}

\int_{\frac{1}{2}}^{1} \frac{4}{9} x (5 - x^{2}) d x

\int_{1}^{t} 39 x^{- 3} d x

What is the derivative of f(x)=sqrt(x)+\sqrt[4]{x} ?
The derivative of f(x)=sqrt(x)+\sqrt[4]{x} is 1/(2sqrt(x))+1/(4x^{3/4)}
What is the first derivative of f(x)=sqrt(x)+\sqrt[4]{x} ?
The first derivative of f(x)=sqrt(x)+\sqrt[4]{x} is 1/(2sqrt(x))+1/(4x^{3/4)}