What is the derivative of 2e^x+8/(\sqrt[3]{x)} ?

The derivative of 2e^x+8/(\sqrt[3]{x)} is 2e^x-8/(3x^{4/3)}

What is the first derivative of 2e^x+8/(\sqrt[3]{x)} ?

The first derivative of 2e^x+8/(\sqrt[3]{x)} is 2e^x-8/(3x^{4/3)}

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derivative of

2 e^{x} - \frac{8}{3 x ^{\frac{4}{3}}}

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (2 e^{x}) = 2 e^{x}

= 2 e^{x} - \frac{8}{3 x ^{\frac{4}{3}}}

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

x y - 2 y^{2} = 5, (11, 5)

\int_{1}^{2} \frac{9}{x ^{3} + 4 x} d x

\frac{d}{d x} (\frac{5}{2} x^{2})

y = - \frac{6}{x}, at x = - 4

What is the derivative of 2e^x+8/(\sqrt[3]{x)} ?
The derivative of 2e^x+8/(\sqrt[3]{x)} is 2e^x-8/(3x^{4/3)}
What is the first derivative of 2e^x+8/(\sqrt[3]{x)} ?
The first derivative of 2e^x+8/(\sqrt[3]{x)} is 2e^x-8/(3x^{4/3)}