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seems like you want to take a derivative to me
I think thats what they're asking, so I would derive that equation to get 8x
almost, y' = 8xx'
do you know what an implicit is?
or rather, just take the derivative of both sides with respect to time, not with respect to x
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Implicit differentiation?
yes \[y = 4x^2+1\] \[\frac{d}{dt}y = \frac{d}{dt}4x^2+\frac{d}{dt}1\] \[\frac{dy}{dt} = \frac{dx}{dt}8x+0\]
dx/dt has been defined for us as 2
So now I would plug in my values?
of course
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So it is increasing at a rate of 16?
correct
the key is seeing that x' is not defined as dx/dx in this case, so we take a derivative implicitly with respect to time, and not to x
Okay ill remember that for next time, thank you so much for your help!!!!!!
good luck
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