Question 1160094
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Another way....

{{{ f(x) = (4x-5)^3*(5x^2+8)^3 }}}

Use the product & chain rules:

Product rule  d(f(x)g(x))/dx = f(x)g'(x) + f'(x)g(x)

Example:  f'( {{{ (3x+2)(2x^2+1) }}}) =  {{{ (3x+2)(4x) + (2x^2+1)(3) }}} and then simplify if necessary.  Yes, that simple!  

"First times the derivative of the 2nd, plus the 2nd times the derivative of the first."



Chain rule    (f(g(x)))' = f'(g(x))*g'(x)   

Example:  f' ( {{{ (3x^2+2)^4 }}} ) = {{{ 4(3x^2+2)^3 * (6x) }}} 

"Derivative of the expression times the derivative of what's inside."


f'(x) = {{{ (4x-5)^3* ( 3*(5x^2+8)^2* 10x ) + (5x^2+8)^3*  ( 3*(4x-5)^2 *4 ) }}}

Factor out {{{ (4x-5)^2*(5x^2+8)^2 }}}
 = {{{ (4x-5)^2*(5x^2+8)^2 * (180x^2-150x+96) }}} 
 = {{{ highlight (6*(4x-5)^2*(5x^2+8)^2 * (30x^2-25x+16) ) }}}