#\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}}}#
