Question

let f be a differentiable function such that f(1)=2 and f'(x)=f(x) for all x ∈ R.if h(x)=f(f(x)) then dh/dx at x=1 is

Answer 1

f is differentiable function such that, f(1) = 2 and f'(x) = f(x)

f'(x) = f(x)

⇒f'(x)/f(x) = 1

⇒∫f'(x)/f(x) dx = ∫dx

⇒lnf(x) = x + c, where c is constant

it is given, f(1) = 2

so, lnf(1) = 1 + c

⇒ln2 = 1 + c

⇒c = ln2 - 1

hence, lnf(x) = x + ln2 - 1

⇒lnf(x) - ln2 = (x - 1)

⇒ln [f(x)/2 ] = (x - 1)

⇒f(x)/2 = e^(x - 1)

⇒f(x) = 2e^(x - 1)

now, h(x) = f(f(x))

= f(2e^(x - 1))

= 2e^{2e^(x - 1) -1}

now, dh(x)/dx = 2e^{2e^(x - 1)-1} × 2e^(x - 1)

at x = 1,

dh/dx = 2e^{2e^0 - 1} × 2e^(1 - 1)

= 4e

hence, answer is 4e