Linear Approx.

**Linnus** · October 21st 2007, 07:15 PM

I know how to do "normal" linear approx...but I don't know how to diff this equation. F(x)= 2^x^2 and approx f(.9) Thank you!

**Jhevon** · October 21st 2007, 07:21 PM

Originally Posted by Linnus

I know how to do "normal" linear approx...but I don't know how to diff this equation. F(x)= 2^x^2 and approx f(.9) Thank you!

by the chain rule: (or you could notice that $2^{x^2} = e^{x^2 \ln 2}$ )

$\frac d{dx}2^{x^2} = 2x \ln 2 \cdot 2^{x^2} = x \ln 2 \cdot 2^{x^2 + 1}$

the tangent line approximation is given by:

$f(x) \approx f(a) + f'(a)(x - a)$

where $x = 0.9$ and $a = 1$

by "normal" linear approx, i guess you want the normal line. this is just the line perpendicular to the tangent line

**Linnus** · October 21st 2007, 07:32 PM

Thanks, just disregard the normal thing

I just want to confirm is this is right..
f(.9)=2+ln2 (4) (-.1)=2-.4ln2

Thanks!

**Jhevon** · October 21st 2007, 07:38 PM

Originally Posted by Linnus

Thanks, just disregard the normal thing

I just want to confirm is this is right..
f(.9)=2+ln2 (4) (-.1)=2-.4ln2

Thanks!

correct

the actual answer: 1.7532114...

our approximation: 1.72274...

not that bad, but i've seen better, we're only good up to one decimal place here

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