solve this one

If for a continous function f,f(0)=f(1)=0 and f'(1)=2 and g(x)=f(e^x).e^(f(x)) then g'(0)=___

1 Answers

well,msp, g'(x) = f'(e^x) . e^x . e^f(x) + e^f(x) . f'(x) . f(e^x)

now,putting x=0 , g'(0) = 2.1.1 + 1.0.1 = 2

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