given dy/dx[f(2x)]=f'(x) and f'(1)=1, find f'(2)
@Hero and @FibonacciChick666 please help
@thomaster
ok, so can you tell me what dy/dx stands for?
also, do you have more information?
No more info, and dy/dx is the change in y in respect to the change in x
ok hmm, this is interesting, I have a theory for the problem, but I can't guarantee it
integrate both sides wrt x , what do you get?
or WAIT easier
can you re-write f'(x)?
in general
I tried thinking that since they said \[\Large \frac{ dy }{ dx }\left[ f(2x) \right]=f \prime (x)\]
what did you get?
Isn't it \[\Large f \prime (2x)=f \prime (x) \]
or was the answer wrong?
yea, that's what I'm thinking
ok and here are hte answer choices sorry I should've included those
1/4 1/2 3/4 3/2 5/2
what? that's weird
exactly I'd assume that there would be 1 as an answer
uhm , I have to go but, let's see. we'd need dy/dx f(4)=f'(2)
so maybe something where you split upthe four into f(1)+f(1)+f(1) but I'm not sure
Hmm, ok thanks anyways
this doesn't make sense to me, @ganeshie8 , any ideas?
\[\Large\rm \frac{d}{dx}f(2x)=f'(x)\]Apply chain rule to the left side,\[\Large\rm 2f'(2x)=f'(x)\]Evaluate this at x=1,\[\Large\rm 2f'(2)=f'(1)\]Understand how to wrap it up? :)
That's so cool!
Ya it's a neat problem! I remember seeing this one like a month ago or something like that.
So it'd be 1/2 since the LHS is 1 and you divide by 2 on both sides to just get f'(2)
Yah looks good!
@FibonacciChick666 yeahhh
oh wow I overcomplicated. Sorry, but yea, I can see that working now, as a side note though, that means that every derivative should have the same value since 2x changes to just 2, so you can't get 1 for f'(1) but anyways, go with what zepdrix said becasue it's a better lead than I had
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