Question 1160095

(a) 

Let {{{f(x)= sqrt(9+6x^2)}}} 

Find {{{f}}} ' {{{(x)}}}:


apply the chain rule:{{{(df*u)/(dx)=(df/du)(du/dx)}}}

{{{f=sqrt(u)}}},  {{{u=9+6x^2}}}


 {{{f}}}'{{{(x)=(d/du)(u) *(d/dx)(9+6x^2)}}}


{{{(d/du)(u)=1/(2sqrt(u))}}}

{{{(d/dx)(9+6x^2)=12x}}}



{{{f}}}'{{{(x)=(1/(2sqrt(u)))*12x}}}


substitute back {{{u=9+6x^2}}}

{{{f }}}' {{{(x)=(1/(2sqrt(9+6x^2)))*12x}}}


{{{f}}}'{{{(x)=12x/(2sqrt(9+6x^2))}}}


{{{f}}}'{{{(x)=6x/(sqrt(9+6x^2))}}}



(b) 

Let {{{f(x)= e^sqrt(9+6x^2) }}}

Find {{{f}}}'{{{(x)}}}


apply the chain rule:{{{(df*u)/(dx)=(df/du)(du/dx)}}}

{{{f=e^u}}},  {{{u=sqrt(9+6x^2)}}}

 
{{{f}}}'{{{(x)=(d/du)(e^u) *(d/dx)(sqrt(9+6x^2))}}}


{{{(d/du)(e^u) =e^u}}}

{{{(d/dx)(sqrt(9+6x^2))=6x/sqrt(9+6x^2)}}}


 {{{f}}}' {{{(x)=e^u*(6x/sqrt(9+6x^2))}}}..........substitute back {{{u=sqrt(9+6x^2)}}}


{{{f}}}'{{{matrix(2,1,"",(x)=e^sqrt(9+6x^2)*(6x/sqrt(9+6x^2)))}}}

{{{f}}}'{{{matrix(2,1,"",(x)=(6x*e^sqrt(9+6x^2))/sqrt(9+6x^2)))}}}