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

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

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