Question 1186231

a. {{{f(x) = 1/(2x) - 3}}}

{{{f(x) = y}}}

{{{y = 1/(2x) - 3}}}...........swap variables

{{{x= 1/(2y) - 3}}}.........solve for {{{y}}}

{{{x+3= 1/(2y) }}}

{{{2y= 1/(x+3) }}}

{{{y= 1/(2(x+3)) }}}


=> {{{f}}}'{{{(x) = 1/(2(x+3)) }}}



b. {{{f(x) = 4x^2 - 5}}}

{{{y= 4x^2 - 5}}}

{{{x= 4y^2 - 5}}}

{{{x+5= 4y^2 }}}

{{{(x+5)/4= y^2 }}}

{{{y}}}= ± {{{sqrt((x+5)/4) }}}

 {{{f}}}'{{{(x)}}}= ± {{{sqrt(x + 5)/2}}}



c. {{{A(r) = 4(Pi)r^2}}}

{{{r = 4(Pi)(A(r))^2}}}.........swap variables 

{{{r/4(Pi)=(A(r))^2}}}

{{{A}}}'{{{(r)}}}= ± {{{sqrt(r)/(2sqrt(pi))}}}