Find `f'' (1) if f(x) = 2(2^(x))`

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lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

`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` .

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

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

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

`ln(f(x))=ln(2)+xln(2)`      (i)

Differentiate (i) with respect to x ,

`(1/(f(x)))f'(x)=ln(2)`

`f'(x)=ln(2)f(x)`(ii)

Differentiate (ii) with respect to x again,

`f''(x)=ln(2)f'(x)`             (iii)

Substitute value of f'(x) from (ii) in (iii), we have

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

`f''(1)=(ln(2))^2f(1)`

`=(ln(2))^2(2xx2)`

`=(2ln(2))^2`

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