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}}}