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

The derivative of sqrt(x)+1/(\sqrt[3]{x^4}) is 1/(2sqrt(x))-4/(3x^{7/3)}

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

The first derivative of sqrt(x)+1/(\sqrt[3]{x^4}) is 1/(2sqrt(x))-4/(3x^{7/3)}

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Popular >

\frac{1}{2 x} - \frac{4}{3 x ^{\frac{7}{3}}}

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{4}{3 x ^{\frac{7}{3}}}

\frac{\partial}{\partial x} (- e^{- \frac{y}{x}})

\int \frac{3 ^{l n (x)}}{x} d x

\frac{\partial}{\partial u} (v^{2})

\int \frac{1}{( x ^{2} ) ^{2}} d (x^{2})^{2}

\frac{\partial}{\partial x} (2 x^{4} y + cos (x^{3}))

What is the derivative of sqrt(x)+1/(\sqrt[3]{x^4}) ?
The derivative of sqrt(x)+1/(\sqrt[3]{x^4}) is 1/(2sqrt(x))-4/(3x^{7/3)}
What is the first derivative of sqrt(x)+1/(\sqrt[3]{x^4}) ?
The first derivative of sqrt(x)+1/(\sqrt[3]{x^4}) is 1/(2sqrt(x))-4/(3x^{7/3)}