## mathcalculus Group Title 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 9 months ago 9 months ago

1. mathcalculus Group Title

why are my answers wrong? :(

2. phi Group Title

for part a) you should have dy =16xh+31h

3. mathcalculus Group Title

why have 31 h?

4. phi Group Title

if you start with y= x d (y = x) is dy = dx or using h instead of dx dy = h for your problem $d (y = 8 x^2+ 31 x + 12 )\\ dy = 8 dx^2 + 31 dx \\ dy = 16x dx +31 dx \\ dy = 16x \ h + 31 \ h$

5. mathcalculus Group Title

i thought we leave the constant alone?

6. mathcalculus Group Title

oh because it was an x because finding the derivative?

7. phi Group Title

compared with $d (y = 31 x ) \\ dy = 31\ dx$

8. mathcalculus Group Title

yes it ='s to 1

9. phi Group Title

$\frac{d}{dx} (y = 31 x) \\ \frac{d}{dx}y = 31 \frac{d}{dx}x\\ \frac{dy}{dx}= 31\frac{dx}{dx}\\ \frac{dy}{dx}= 31\cdot 1=31$

10. mathcalculus Group Title

right.

11. mathcalculus Group Title

but i was confused why i couldn't leave 31 alone

12. mathcalculus Group Title

13. phi Group Title

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 $\Delta y = \frac{dy}{dx} \Delta x =(16x + 31)\Delta x= (16x + 31)h$

14. mathcalculus Group Title

oh okay, got cha.

15. mathcalculus Group Title

what if it had like an x and a y inside

16. mathcalculus Group Title

could i still apply this?

17. phi Group Title

it would be the same idea, but you use "partial derivatives"

18. mathcalculus Group Title

ex: dy/dx

19. mathcalculus Group Title

so (d/dx+dy/dx)h ?

20. phi Group Title

do you have a specific question?

21. phi Group Title

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

22. mathcalculus Group Title

well i know that we have to do this: 2x*y(dy/dx) +2y*(d/dx).. right?

23. phi Group Title

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 $\Delta z = 2x \Delta x + 2y \Delta y$

24. phi Group Title

But you won't see those type of problems.

25. mathcalculus Group Title

oh okay, thank you so much! :)