Question

Finding formula for differential dy. Please check.

Answer 1

One thing...

Answer 2

The derivative is dy/dx, (change in y divided by change in x, or, the slope).
Not just change in y.

Answer 3

So you would have dy/dx = ...
in the first row

And then multiply both sides by dx to obtain the formula for dy.

Answer 4

which part are you under? b?

Answer 5

I am referring to part a.

Answer 6

ooh so I skipped a step

Answer 7

no, just left out the dx.
(I explained how and what)

Answer 8

like this?

Answer 9

Beautiful

Answer 10

Ok, now part two...
Just want to verify that f is:

\(\large\color{black}{\displaystyle f(x)=\sqrt[4]{1+3x}}\)

Answer 11

Finding linear approx for a=5, is just the same as finding the tangent line to the curve at x=5.

Answer 12

I don't think you are approximating f(5), because you get an integer output for f(5) anyway.... rather, (I think that) you need to find a linear approximation function or the tangent line for a=5, which can be used to approximate the function for values near 5.

Answer 13

mm well the formula is
y=f(a)+f'(a)(x-a)

Answer 14

so then y=f(5)+f'(5)(x-5)

Answer 15

maybe I am wrong, but it is reasonable to think that you need a tangent line at x=5 (or linearization of the function for values near 5), rather than just f(5).

Answer 16

And yes, y=f(5)+f'(5)(x-5) is correct.

Answer 17

http://tutorial.math.lamar.edu/Classes/CalcI/LinearApproximations.aspx

Answer 18

right, so the third and fourth lines under part b are figuring out f(5) and f'(5)
and then after that plugging it into the equation.

Answer 19

\(\large\color{black}{\displaystyle f(x)=\sqrt[4]{1+3x}}\)
\(\large\color{black}{\displaystyle f(5)=\sqrt[4]{1+3\cdot5}=2}\)



\(\large\color{black}{\displaystyle f'(x)=\frac{1}{4\sqrt[4]{(1+3x)^3}}}\)

\(\large\color{black}{\displaystyle f'(5)=\frac{1}{4\sqrt[4]{(1+3\cdot 5)^3}}=\frac{1}{4\times 2^3}=1/32}\)


(If I didn't make simple algebraic erros)

Answer 20

It seems reasonable, without having any error that is obvious right off, to me... bt checking is never bad.

Answer 21

Yeah that looks right. So it would just be better in fraction form?

Answer 22

What do you mean?

Answer 23

I already got both those numbers :)
only I wrote 1/32 as 0.03125

Answer 24

Yeah, that confused me at first just a bit... try to use fractions if decimals look like some abstruse nonsense.

Answer 25

\(\large\color{black}{\displaystyle L(x)=2+\frac{1}{32}(x-5)}\)

Answer 26

I denoted it as L because it stands for linearization...
not that this matters.

Answer 27

Ah ok, this all make sense ^-^

Answer 28

so if I make those changes to b, part c is all good still?
[after replacing the decimal numbers with 1/32]

Answer 29

Yes, use 1/32, it is more convinient and exact... but I don't see major errors in either case.

Just that fractions are benevolent, and decimals are evil demons.

Answer 30

hahahah

Answer 31

YOu did something irrelevant, don't get me wrong...

Answer 32

You need to approximate \(\sqrt[4]{15.97}\), (or f(15.97) )
You will use your tangent line to approximate that.

Answer 33

That is what the instructions say, although the result would likely be extermely innaccurate.

Answer 34

hm? so if you do it that way what would the result be?
In class this is how the instructor did it.
Except i wrote
f(x)=2+1/32(4.99-5)
when it should be
f(4.99)=2+1/32(4.99-5)

Answer 35

\(\large\color{black}{\displaystyle L(x)=2+\frac{1}{32}(x-5)}\)

\(\large\color{black}{\displaystyle L(15.97)=2+\frac{1}{32}(15.97-5)}\)

Answer 36

I might be wrong, but that is my best take on it...

Answer 37

No no, because we want it to equal
sqroot (15.97)
you're leaving out the 1+3x under there

Answer 38

You're doing: sqroot(1+3x+15.97) if that makes sense

Answer 39

true:

Answer 40

What you did is right, and I made an error.

Answer 41

f(x)=2+1/32(4.99-5) is right

Answer 42

Yay! good because I wasnt sure haha

Answer 43

So..now it's all good? xP

Answer 44

I suppose, so :)

GOOD LUCK !

Answer 45

I hope so haha, we'll see.

Thanks you lovely human, you!

Answer 46

yw