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}}}}
f ( x ) = x 2 1 − x 2 2 {\displaystyle f(x)=x^{2}{\sqrt[{2}]{1-x^{2}}}} simplify
f ( x ) = x 2 ( 1 − x 2 ) 1 2 {\displaystyle f(x)=x^{2}{(1-x^{2})}^{1 \over 2}}
f ′ ( x ) = x 2 ( 1 2 ( 1 − x 2 ) − 1 2 ( − 2 x ) ) + 2 x ( 1 − x 2 ) 1 2 {\displaystyle f'(x)=x^{2}(1 \over 2{(1-x^{2})}^{-1 \over 2}(-2x))+2x{(1-x^{2})}^{1 \over 2}}
f ′ ( x ) = − x 3 ( 1 − x 2 ) − 1 2 + 2 x ( 1 − x 2 ) 1 2 {\displaystyle f'(x)=-x^{3}{(1-x^{2})}^{-1 \over 2}+2x{(1-x^{2})}^{1 \over 2}}
f ′ ( x ) = x ( 1 − x 2 ) − 1 2 [ − x 2 + 2 ( 1 − x 2 ) ] {\displaystyle f'(x)=x{(1-x^{2})}^{-1 \over 2}[-x^{2}+2{(1-x^{2})}]}
f ′ ( x ) = x ( 1 − x 2 ) 1 2 [ − x 2 + 2 − 2 x 2 ] {\displaystyle f'(x)={x \over {(1-x^{2})}^{1 \over 2}}[{-x^{2}+2-2x^{2}}]}
f ′ ( x ) = x ( 1 − x 2 ) 1 2 [ 2 − 3 x 2 ] {\displaystyle f'(x)={x \over {(1-x^{2})}^{1 \over 2}}{[2-3x^{2}]}}
f ′ ( x ) = x [ 2 − 3 x 2 ] ( 1 − x 2 ) 1 2 {\displaystyle f'(x)={x[2-3x^{2}] \over {(1-x^{2})}^{1 \over 2}}}
f ′ ( x ) = [ 2 x − 3 x 3 ] ( 1 − x 2 ) 1 2 {\displaystyle f'(x)={[2x-3x^{3}] \over {(1-x^{2})}^{1 \over 2}}}
f ( x ) = ( x + 3 ) 4 ( x 2 + 5 ) 1 2 {\displaystyle f(x)={(x+3)^{4} \over (x^{2}+5)^{1 \over 2}}}
The Quotient Rule ( f g ) ′ = ( g f ′ − f g ′ ) g 2 {\displaystyle ({f \over g})'={(gf'-fg') \over g^{2}}}
f = ( x + 3 ) 4 {\displaystyle f={(x+3)}^{4}}
f ′ = 4 ( x + 3 ) 3 {\displaystyle f'=4(x+3)^{3}}
g = ( x 2 + 5 ) 1 2 {\displaystyle g=(x^{2}+5)^{1 \over 2}}
g ′ = 1 2 ( x 2 + 5 ) − 1 2 ( 2 x ) {\displaystyle g'={1 \over 2}(x^{2}+5)^{-1 \over 2}(2x)}
( x 2 + 5 ) 1 2 4 ( x + 3 ) 3 + ( x + 3 ) 4 1 2 ( x 2 + 5 ) − 1 2 ( 2 x ) ( x 2 + 5 ) {\displaystyle {{(x^{2}+5)^{1 \over 2}4(x+3)^{3}+{(x+3)}^{4}{1 \over 2}(x^{2}+5)^{-1 \over 2}(2x)} \over (x^{2}+5)}}
4 ( x 2 + 5 ) ( x + 3 ) 3 − x ( x + 3 ) 4 ( x 2 + 5 ) − 1 2 ( x 2 + 5 ) {\displaystyle {{4{(x^{2}+5)}(x+3)^{3}-x{(x+3)}^{4}(x^{2}+5)^{-1 \over 2}} \over (x^{2}+5)}}
( x 2 + 5 ) − 1 2 ( x + 3 ) 3 [ 4 ( x 2 + 5 ) − x ( x + 3 ) ] ( x 2 + 5 ) {\displaystyle (x^{2}+5)^{-1 \over 2}{(x+3)}^{3}[4(x^{2}+5)-x(x+3)] \over (x^{2}+5)}
( x + 3 ) 3 [ 4 ( x 2 + 5 ) − x ( x + 3 ) ] ( x 2 + 5 ) ( x 2 + 5 ) 1 2 {\displaystyle {(x+3)}^{3}[4(x^{2}+5)-x(x+3)] \over (x^{2}+5)(x^{2}+5)^{1 \over 2}}
( x + 3 ) 3 [ 4 x 2 + 20 − x 2 − 3 x ] ( x 2 + 5 ) ( x 2 + 5 ) 1 2 {\displaystyle {(x+3)}^{3}[4x^{2}+20-x^{2}-3x] \over (x^{2}+5)(x^{2}+5)^{1 \over 2}}