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mathcalculus
 one year ago
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^231x
II. Evaluate each of these quantities if x=10 and h=0.04
a. dy= 37.4
b. \Delta y = 7.6528
mathcalculus
 one year ago
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^231x 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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mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1why are my answers wrong? :(

phi
 one year ago
Best ResponseYou've already chosen the best response.1for part a) you should have dy =16xh+31h

phi
 one year ago
Best ResponseYou've already chosen the best response.1if you start with y= x d (y = x) is dy = dx or using h instead of dx dy = h for your problem \[ d (y = 8 x^2+ 31 x + 12 )\\ dy = 8 dx^2 + 31 dx \\ dy = 16x dx +31 dx \\ dy = 16x \ h + 31 \ h \]

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1i thought we leave the constant alone?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1oh because it was an x because finding the derivative?

phi
 one year ago
Best ResponseYou've already chosen the best response.1compared with \[ d (y = 31 x ) \\ dy = 31\ dx \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{d}{dx} (y = 31 x) \\ \frac{d}{dx}y = 31 \frac{d}{dx}x\\ \frac{dy}{dx}= 31\frac{dx}{dx}\\ \frac{dy}{dx}= 31\cdot 1=31 \]

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1but i was confused why i couldn't leave 31 alone

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1and why i had to add h in the end..

phi
 one year ago
Best ResponseYou've already chosen the best response.1maybe 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 \[ \Delta y = \frac{dy}{dx} \Delta x =(16x + 31)\Delta x= (16x + 31)h \]

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1oh okay, got cha.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1what if it had like an x and a y inside

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1could i still apply this?

phi
 one year ago
Best ResponseYou've already chosen the best response.1it would be the same idea, but you use "partial derivatives"

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1so (d/dx+dy/dx)h ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1do you have a specific question?

phi
 one year ago
Best ResponseYou've already chosen the best response.1it 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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1well i know that we have to do this: 2x*y(dy/dx) +2y*(d/dx).. right?

phi
 one year ago
Best ResponseYou've already chosen the best response.1are 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
 one year ago
Best ResponseYou've already chosen the best response.1But you won't see those type of problems.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1oh okay, thank you so much! :)
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