What is the derivative of y=6e^x+8/(\sqrt[3]{x)} ?

The derivative of y=6e^x+8/(\sqrt[3]{x)} is 6e^x-8/(3x^{4/3)}

What is the first derivative of y=6e^x+8/(\sqrt[3]{x)} ?

The first derivative of y=6e^x+8/(\sqrt[3]{x)} is 6e^x-8/(3x^{4/3)}

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derivative of

6 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} (6 e^{x}) = 6 e^{x}

= 6 e^{x} - \frac{8}{3 x ^{\frac{4}{3}}}

\int (x^{2} ln (2 x)) d x

e^{x^{2}} - 1

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

y^{^{'}} = \frac{y}{x}

\frac{\partial}{\partial y} (\frac{1}{ln ( 10 ) ( x - y )})

What is the derivative of y=6e^x+8/(\sqrt[3]{x)} ?
The derivative of y=6e^x+8/(\sqrt[3]{x)} is 6e^x-8/(3x^{4/3)}
What is the first derivative of y=6e^x+8/(\sqrt[3]{x)} ?
The first derivative of y=6e^x+8/(\sqrt[3]{x)} is 6e^x-8/(3x^{4/3)}