Simplifying Derivatives

example 1

$a'(x)=x^{3}{(2x-5)}^{4}$

Product Rule $(f*g)'=f*g'+g*f'$

$f=x^{3}$ $f'=3x^{2}$ $g={(2x-5)}^{4}$ $g'=4{(2x-5)}^{3}(2)$

$a'(x)=(3x^{2})({(2x-5)}^{4})+(x^{3})(4{(2x-5)}^{3}(2))$

$a'(x)=(3x^{2})({(2x-5)}^{4})+(x^{3})(8{(2x-5)}^{3})$

factorize

$a'(x)=x^{2}({(2x-5)}^{3})(8x)+3(2x-5))$

$a'(x)=x^{2}({(2x-5)}^{3})(8x+6x-15))$

$a'(x)=x^{2}{(2x-5)}^{3}(14x-15))$

example 2

$f'(x)=x^{2}{\sqrt[{2}]{4-9x}}$ simplify

Chain rule $[f(g(x))]'=f'(g(x))*g'(x)$

$2x{\sqrt[{2}]{4-9x}}+{-9 \over 2}{(4-9x)}^{-1 \over 2}$

${x(8-{45x \over 2}){\sqrt[{2}]{4-9x}} \over {\sqrt[{2}]{4-9x}}}$

example 3

$f(x)=x^{2}{\sqrt[{2}]{1-x^{2}}}$ simplify

$f(x)=x^{2}{(1-x^{2})}^{1 \over 2}$

Product Rule $(f*g)'=f*g'+g*f'$

$f'(x)=x^{2}(1 \over 2{(1-x^{2})}^{-1 \over 2}(-2x))+2x{(1-x^{2})}^{1 \over 2}$

$f'(x)=-x^{3}{(1-x^{2})}^{-1 \over 2}+2x{(1-x^{2})}^{1 \over 2}$

factorize

$f'(x)=x{(1-x^{2})}^{-1 \over 2}[-x^{2}+2{(1-x^{2})}]$

Failed to parse (syntax error): {\displaystyle f'(x)= {x \over {(1-x^2)}^{1 \over 2}} [-x^2 + 2 -2x^2}]}

Failed to parse (syntax error): {\displaystyle f'(x)= {x \over {(1-x^2)}^{1 \over 2}} [ 2 - 3x^2}]}

Failed to parse (syntax error): {\displaystyle f'(x)= {x [ 2 - 3x^2}] \over {(1-x^2)}^{1 \over 2}} }

Failed to parse (syntax error): {\displaystyle f'(x)= {[ 2x - 3x^3}] \over {(1-x^2)}^{1 \over 2}} }