How do you find f'(x) using the definition of a derivative #y=6e^x+4/root3x#?

1 Answer
Feb 23, 2016

#6e^x-4/(3x^(4/3#

Explanation:

#\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}}}#