What is the derivative of \sqrt[5]{x^4}-4/x ?

The derivative of \sqrt[5]{x^4}-4/x is (4(x^{9/5}+5))/(5x^{9/5)\sqrt[5]{x}}

What is the first derivative of \sqrt[5]{x^4}-4/x ?

The first derivative of \sqrt[5]{x^4}-4/x is (4(x^{9/5}+5))/(5x^{9/5)\sqrt[5]{x}}

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Solution steps

Apply the Sum/Difference Rule:

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

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

= \frac{4}{5 x ^{\frac{1}{5}}} - (- \frac{4}{x ^{2}})

Simplify

x \to 0 + lim (tan (2 x))

(8 x + 9)^{- 2}

\frac{d}{d x} (x^{3} - 12 x)

\int \frac{8 ( x ^{2} + 4 )}{x ( x ^{2} + 8 )} d x

\frac{d}{d x} (5 x^{3} + 3 x^{2} - 4 x + 1)

What is the derivative of \sqrt[5]{x^4}-4/x ?
The derivative of \sqrt[5]{x^4}-4/x is (4(x^{9/5}+5))/(5x^{9/5)\sqrt[5]{x}}
What is the first derivative of \sqrt[5]{x^4}-4/x ?
The first derivative of \sqrt[5]{x^4}-4/x is (4(x^{9/5}+5))/(5x^{9/5)\sqrt[5]{x}}