y = 1 + x 2 3 x {\displaystyle y={\frac {1+x^{2}}{3x}}}
Split it into two parts y = 1 3 x + x 3 {\displaystyle y={\frac {1}{3x}}+{\frac {x}{3}}}
Using the sum rule of derivative, ( f + g ) ′ = f ′ + g ′ {\displaystyle (f+g)'=f'+g'} for all functions f and g.
calculate derivative for two parts y ′ = ( 1 3 x ) ′ + ( x 3 ) ′ {\displaystyle y'=({\frac {1}{3x}})'+({\frac {x}{3}})'}
y ′ = 1 3 ( x − 1 ) ′ + 1 3 ∗ 1 {\displaystyle y'={\frac {1}{3}}(x^{-1})'+{\frac {1}{3}}*1}
y ′ = 1 3 ( − x − 2 ) + 1 3 {\displaystyle y'={\frac {1}{3}}(-x^{-2})+{\frac {1}{3}}}
y ′ = − 1 3 x 2 + 1 3 {\displaystyle y'=-{\frac {1}{3x^{2}}}+{\frac {1}{3}}}
answer: − 1 3 x 2 + 1 3 {\displaystyle \color {blue}-{\frac {1}{3x^{2}}}+{\frac {1}{3}}}
for all functions f and g.
