Question 1191921

{{{y=-8x*ln(5x+7)}}}

{{{dy/dx}}}= ?

Differentiate separately both sides of the equation (treat {{{y}}} as a function of {{{x}}}):


{{{dy/dx= (d/dx)( -8x*ln(5x+7))}}}

Differentiate the RHS of the equation. Apply the constant multiple rule.


{{{(d/dx)( -8x*ln(5x+7))}}}


={{{-8(d/dx)( x*ln(5x+7))}}}...........Apply the product rule


={{{-8((d/dx)( x)*ln(5x+7)+x(d/dx)(ln(5x+7)))}}}


={{{-8(d/dx)( x)*ln(5x+7)-8*x(d/dx)(ln(5x+7))}}}


The function {{{ln(5x+7)}}} is the composition {{{f(g(x)) }}}of two functions {{{f(u)=ln(u) }}}and {{{g(x)=5x+7}}}, so


={{{-8x( d/du)(ln(u)) (d/dx)(5x+7))-8ln(5x+7) (d/dx)(x)}}}......The derivative of the natural logarithm is {{{ (d/du)(ln(u))= 1/u}}}

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 ={{{-8x( 1/u) (d/dx)(5x+7)-8ln(5x+7) (d/dx)(x)}}}

={{{(- 8x(d/dx)(5x+7))/u-8ln(5x+7) (d/dx)(x)}}}


Return to the old variable:


={{{(- 8x(d/dx)(5x+7))/(5x+7)-8ln(5x+7) (d/dx)(x)}}}


The derivative of a sum/difference is the sum/difference of derivatives:


={{{(- 8x(d/dx)(5x)+(d/dx)(7))/(5x+7)-8ln(5x+7) (d/dx)(x)}}}


The derivative of a constant is {{{0}}}:{{{(d/dx)(7)=0}}}


={{{(- 8x(d/dx)(5x)+0))/(5x+7)-8ln(5x+7) (d/dx)(x)}}}


Apply the constant multiple rule:{{{(d/dx)(5x)=5(d/dx)(x)=5*1=5}}}


={{{(- 8x(5))/(5x+7)-8ln(5x+7)*1}}}


={{{(- 40x)/(5x+7)-8ln(5x+7) }}}


= {{{-(40x+8(5x+7)ln(5x+7))/(5x+7)}}}


Thus, {{{(d/dx)(-8x*ln(5x+7))=-(40x+8(5x+7)ln(5x+7))/(5x+7)}}}


Therefore, {{{(dy/dx)=-(40x+8(5x+7)ln(5x+7))/(5x+7)}}}

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