What is the derivative of \sqrt[5]{x^2}-x^pi ?

The derivative of \sqrt[5]{x^2}-x^pi is (2x)/(5(x^2)^{4/5)}-pix^{pi-1}

What is the first derivative of \sqrt[5]{x^2}-x^pi ?

The first derivative of \sqrt[5]{x^2}-x^pi is (2x)/(5(x^2)^{4/5)}-pix^{pi-1}

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derivative of

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

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (x^{π}) = π x^{π - 1}

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

\int 2 + t d t

x \to \infty lim (x^{- 2} e^{2 x})

\frac{d}{d x} (e^{(2 x)^{3} - 2})

y^{^{'}} = \frac{( 4 t ^{3} + 1 )}{2 y - 6}

What is the derivative of \sqrt[5]{x^2}-x^pi ?
The derivative of \sqrt[5]{x^2}-x^pi is (2x)/(5(x^2)^{4/5)}-pix^{pi-1}
What is the first derivative of \sqrt[5]{x^2}-x^pi ?
The first derivative of \sqrt[5]{x^2}-x^pi is (2x)/(5(x^2)^{4/5)}-pix^{pi-1}