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HELP :( LINEARIZATION: Please use h in place of \Delta x and dx I. If y= f(x)= 8 x^2+ 31 x + 12 find a. dy =16xh+31 b. \Delta y = 8*(x+h)^2+31*(x+h)-8x^2-31x II. Evaluate each of these quantities if x=10 and h=0.04 a. dy= 37.4 b. \Delta y = 7.6528

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why are my answers wrong? :(
  • phi
for part a) you should have dy =16xh+31h
why have 31 h?

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Other answers:

  • phi
if you start with y= x d (y = x) is dy = dx or using h instead of dx dy = h for your problem \[ d (y = 8 x^2+ 31 x + 12 )\\ dy = 8 dx^2 + 31 dx \\ dy = 16x dx +31 dx \\ dy = 16x \ h + 31 \ h \]
i thought we leave the constant alone?
oh because it was an x because finding the derivative?
  • phi
compared with \[ d (y = 31 x ) \\ dy = 31\ dx \]
yes it ='s to 1
  • phi
\[ \frac{d}{dx} (y = 31 x) \\ \frac{d}{dx}y = 31 \frac{d}{dx}x\\ \frac{dy}{dx}= 31\frac{dx}{dx}\\ \frac{dy}{dx}= 31\cdot 1=31 \]
but i was confused why i couldn't leave 31 alone
and why i had to add h in the end..
  • phi
maybe a better way to think of it is \[ \Delta y = \frac{dy}{dx} \Delta x \] where dy/dx is the derivative of your function \[ \frac{dy}{dx}= 16x + 31\] and \[ \Delta y = \frac{dy}{dx} \Delta x =(16x + 31)\Delta x= (16x + 31)h \]
oh okay, got cha.
what if it had like an x and a y inside
could i still apply this?
  • phi
it would be the same idea, but you use "partial derivatives"
ex: dy/dx
so (d/dx+dy/dx)h ?
  • phi
do you have a specific question?
  • phi
it is a new topic to talk about functions of two variables f(x,y) = x^2 + y^2 for example you won't see those until calculus II
well i know that we have to do this: 2x*y(dy/dx) +2y*(d/dx).. right?
  • phi
are you asking about f(x,y) = x^2 + y^2 or z = x^2 + y^2 ? we find the partial derivative with respect to x (that means treat y as a constant) fx (means partial derivative with respect to x) = 2x fy = 2y we then say \[ \Delta z = 2x \Delta x + 2y \Delta y \]
  • phi
But you won't see those type of problems.
oh okay, thank you so much! :)

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