`f(x)=2(2^x)`

First, determine f'(x). Apply the formula `(a^u)'=lna *a^u *u'` .

`f'(x)=(2(2^x))'`

`f'(x)=2* ln2*2^x*x'`

`f'(x)=2*ln2*2^x*1`

`f'(x)=2ln2(2^x)`

Then, take the derivative of it again to get f"(x).

`f"(x)=2ln2*ln2*2^x*x'`

`f"(x)=2*ln2*ln2*2^x*1`

`f"(x)=2(ln2)^2(2^x)`

Now that the second derivative is know, solve for f"(1). So, plug-in x=1.

`f"(1)=2(ln2)^2(2^1)`

`f"(1)=2*(ln2)^2*2`

`f"(1)=4(ln2)^2`

**Hence, `f"(1)=4(ln2)^2` .**