Q. 244.2( 5 Votes )

# A function

Answer :

Given f(x) is differentiable at x = 0 and f(x) ≠ 0

And f(x + y) = f(x)f(y) also f’(0) = 2

To prove: f’(x) = 2f(x)

As we know that,

f’(x) = as f(x+h) = f(x)f(h)

f’(x) = f’(x) = …(1)

As

f(x + y) = f(x)f(y)

put x = y = 0

f(0+0) = f(0)f(0)

f(0) = {f(0)}2

f(0) = 1 { f(x) ≠ 0 ….given}

equation 1 is deduced as-

f’(x) = f’(x) = f’(x) = f(x)f’(0) {using formula of derivative}

f’(x) = 2 f(x) …proved { it is given that f’(0) = 2}

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