If f(x) + x2[f(x)]4 = 18 and f( 1) = 2, find f ‘(1). f'(1) =

f(x) + x2[f(x)]4 = 18

differentiationg on both sides

==> f ‘(x) + { (2x)[f(x)]4 + x2

4[f(x)]3 f ‘(x) } =

0

since (UV)’ = U’V + UV’ ; d/dx xn = nxn-1

==> f ‘(x) + (2x)[f(x)]4 + x2

4[f(x)]3 f ‘(x) = 0

==> f ‘(x) + x2 4[f(x)]3 f ‘(x) +

(2x)[f(x)]4 = 0

==> f ‘(x) + x2 4[f(x)]3 f ‘(x) =

-(2x)[f(x)]4

==> { 1 + x2 4[f(x)]3} f ‘(x) =

-(2x)[f(x)]4

==> f ‘(x) = -(2x)[f(x)]4/ { 1 + x2

4[f(x)]3}

==> f ‘(1) = -(2(1))[f(1)]4/ { 1 + (1)2

4[f(1)]3}

==> f ‘(1) = -(2)[2]4/ { 1 + 4[2]3}

==> f ‘(1) = -32/33