Let E = $( 2 A + B - 3 ) ^ { 2 }$ + (A + B + 2) + (4A + B - 1 $)^2$ . What is $\frac { \partial \mathrm { E } } { \partial \mathrm { A } }$ ?
A) 42A + 24B - 9
B) 14A + 6B - 4
C) 42A + 14B - 9
D) 24A + 6B - 4
E) none of these

Question

Quizplus · Accepted Answer

To find $\frac { \partial \mathrm { E } } { \partial \mathrm { A } }$, we differentiate E with respect to A. Given $E = ( 2 A + B - 3 ) ^ { 2 } + (A + B + 2) + (4A + B - 1)^2$,$\frac { \partial \mathrm { E } } { \partial \mathrm { A } }$ = $2(2A + B - 3)(2) + 1 + 2(4A + B - 1)(4)$= $4(2A + B - 3) + 1 + 8(4A + B - 1)$= $8A + 4B - 12 + 1 + 32A + 8B - 8$= $40A + 12B - 19$None of the given options match this result.

Let E = (2A+B−3)2( 2 A + B - 3 ) ^ { 2 }(2A+B−3)2

Let E = $( 2 A + B - 3 ) ^ { 2 }$