a ′ ( x ) = x 3 ( 2 x − 5 ) 4 {\displaystyle a'(x)=x^{3}{(2x-5)}^{4}}
Product Rule ( f ∗ g ) ′ = f ∗ g ′ + g ∗ f ′ {\displaystyle (f*g)'=f*g'+g*f'}
f = x 3 {\displaystyle f=x^{3}} f ′ = 3 x 2 {\displaystyle f'=3x^{2}} g = ( 2 x − 5 ) 4 {\displaystyle g={(2x-5)}^{4}} g ′ = 4 ( 2 x − 5 ) 3 ( 2 ) {\displaystyle g'=4{(2x-5)}^{3}(2)}
a ′ ( x ) = ( 3 x 2 ) ( ( 2 x − 5 ) 4 ) + ( x 3 ) ( 4 ( 2 x − 5 ) 3 ( 2 ) ) {\displaystyle a'(x)=(3x^{2})({(2x-5)}^{4})+(x^{3})(4{(2x-5)}^{3}(2))}
a ′ ( x ) = ( 3 x 2 ) ( ( 2 x − 5 ) 4 ) + ( x 3 ) ( 8 ( 2 x − 5 ) 3 ) {\displaystyle a'(x)=(3x^{2})({(2x-5)}^{4})+(x^{3})(8{(2x-5)}^{3})} factorize a ′ ( x ) = x 2 ( ( 2 x − 5 ) 3 ) ( 8 x ) + 3 ( 2 x − 5 ) ) {\displaystyle a'(x)=x^{2}({(2x-5)}^{3})(8x)+3(2x-5))}
a ′ ( x ) = x 2 ( ( 2 x − 5 ) 3 ) ( 8 x + 6 x − 15 ) ) {\displaystyle a'(x)=x^{2}({(2x-5)}^{3})(8x+6x-15))} a ′ ( x ) = x 2 ( 2 x − 5 ) 3 ( 14 x − 15 ) ) {\displaystyle a'(x)=x^{2}{(2x-5)}^{3}(14x-15))}
f ′ ( x ) = x 2 4 − 9 x 2 {\displaystyle f'(x)=x^{2}{\sqrt[{2}]{4-9x}}} simplify
Chain rule [ f ( g ( x ) ) ] ′ = f ′ ( g ( x ) ) ∗ g ′ ( x ) {\displaystyle [f(g(x))]'=f'(g(x))*g'(x)}
2 x 4 − 9 x 2 + − 9 2 ( 4 − 9 x ) − 1 2 {\displaystyle 2x{\sqrt[{2}]{4-9x}}+{-9 \over 2}{(4-9x)}^{-1 \over 2}}
x ( 8 − 45 x 2 ) 4 − 9 x 2 4 − 9 x 2 {\displaystyle {x(8-{45x \over 2}){\sqrt[{2}]{4-9x}} \over {\sqrt[{2}]{4-9x}}}}