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Ok, see let f(x) = (2x-pie)3+2x-cosx. Let g(x) be the inverse of f(x). Hence, g ( f(x) ) = x => g'(f(x)) . f'(x) = 1 Now, we have to find g' ( f(x) ) at x = pie. Hence, f(pie) = (pie)3 +2(pie) + 1 (1) f'(x) = 3 (2x - pie)2 .2 + 2 + Sinx f'(pie) = 6(pie)2 + 2 (2)
Putting the values of (1) and (2) in the eqn. g'(f(x)) . f'(x) = 1 ; g' ( ..) = 1/ (6 (pie)2 + 2) = the value of d/dx(inverse f(x)) at x=pie
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