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!
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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 ) the tangent line approximation is given by: where and by "normal" linear approx, i guess you want the normal line. this is just the line perpendicular to the tangent line
Thanks, just disregard the normal thing I just want to confirm is this is right.. f(.9)=2+ln2 (4) (-.1)=2-.4ln2 Thanks!
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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