\frac{d}{dx}(6e^x+\frac{4}{\sqrt[3]{x}})
Applying sum/difference rule: (f\pm g)^'=f^'\pm g^'
=\frac{d}{dx}(6e^x)+\frac{d}{dx}(\frac{4}{\sqrt[3]{x}})
We have,
=\frac{d}{dx}(6e^x) = 6e^x
(Taking the constant out : (a\cdot f)^'=a\cdot f^' i.e =6\frac{d}{dx}(e^x) and applying common derivative of \frac{d}{dx}(e^x)=e^x)
And,
\frac{d}{dx}(\frac{4}{\sqrt[3]{x}})=-\frac{4}{3x^{\frac{4}{3}}}
(Taking the constant out : (a\cdot f)^'=a\cdot f^' i.e =4\frac{d}{dx}(\frac{1}{\sqrt[3]{x}}) and applying power rule for =4\frac{d}{dx}(x^{-\frac{1}{3}}) i.e \frac{d}{dx}(x^a)=a\cdot x^{a-1} we get =-\frac{4}{3x^{\frac{4}{3}}})
So, finally
6e^x - \frac{4}{3x^{\frac{4}{3}}}