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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\frac{d}{d x} (2 e^{x} + \frac{8}{3 x})

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

Solution steps

\frac{d}{d x} (2 e^{x} + \frac{8}{3 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 e^{x}) + \frac{d}{d x} (\frac{8}{3 x})

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

\frac{d}{d x} (\frac{8}{3 x}) = - \frac{8}{3 x ^{\frac{4}{3}}}

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

\frac{d y}{d x} - x y = 1

\int_{0}^{1} \frac{7}{1 - x ^{2}} d x

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

y = 5 sec (5 x)

(5.6) (5.9)

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)}