/sci/ - Science & Math

[Anonymous No.5065490]

Can someone give me a walkthrough or tips on this problem? I've trie it multiple times and still come up short.

[Anonymous No.5065505]

I've tried multiplying by the conjugate and combining it into one whole fraction but I seem to be doing something wrong.

[Anonymous No.5065508]

>>5065490

the tangent is the slope of the derivative at a point on the curve.

so you must defferentiate the function:

y = 7 * x ^ -1/2

which is dy/dx = -7/2x

which is linear, and the slope is -7/2 which is also the tangen.

for part b do the same thing at points specified.

(homework is against the rules, so once you have your answer, please omit this thraed so you don't get a ban)

[Anonymous No.5065531]

>>5065505

that will never work

[Anonymous No.5065544]

>>5065508
That derivative is rrrong. You should have noticed when the answer said the slope is linear when the curve was not parabolic.
For the derivative of that curve, go to wolframalpha.com and tell it to derive that function.
Homework threads are fine with me and if moot doesnt want them here he should make a math homework board.

If you still cant solve it, post what you have tried and we might correct it.

[Anonymous No.5065553]

>>5065490
Since sqrt(x) is equal to x^(1/2), you can mutliply the 7 with its inverse instead of dividing by it, which leads to 7*x^(-1/2). That is about as pretty as you will get it.

[Anonymous No.5065569]

>>5065544

thank you i will try that

is wolfrramalpha free or subscripton?

[Anonymous No.5065607]

>>5065544
wolframalpha.com didn't really help so much but i tried it again and so far im at:
f'(x) = lim as h->0 ((7(x+h)^-1/2 - 7x^-1/2)/h)

[Anonymous No.5065609]

>>5065607
not really sure how that helped finding the derivative, it seemed easier with the fractions...

[Anonymous No.5065625]

>>5065609

it is easier with the fractions.

that guys just trying to troll us.

he probably sells that wolfram book himself, or gets a cut.

[Anonymous No.5065637]

>>5065625
so do you know what i should be doing?

[Anonymous No.5065652]

>>5065607
Oh, you dont need to make it that complicated. Just do
http://www.wolframalpha.com/input/?i=derive+7x%5E-1%2F2
Poster from >>5065544 and >>5065553

[Anonymous No.5065656]

>>5065637

yes i put it up there before that guy started pushing his book.

[Anonymous No.5065662]

>>5065625
>>5065656
wolframalpha is no book and I get no cuts, it just is able to solve >50% of the homework problems posted here

[Anonymous No.5065664]

>>5065652
oh damn yeah i got something completely different haha
though when i looked at the steps, it looks like it doesnt even factor in the f(x+h) - f(x)/h

[Anonymous No.5065671]

>>5065625
I think deriving 7x^(-1/2) is easier, since the derivative of x^n is n*x^(n-1) and therefore the derivative of x^(-1/2) is (-1/2)*x^(-3/2).

[Wolfram Book Guy No.5065686]

>>5065664
d(x^n)/dx
=lim h->0 ((x+h)^n-x^n)/h
=lim h->0 ((x^n+n*h*x^(n-1)+h²*(...))-x^n)/h
=lim h->0 (n*h*x^(n-1)+h²*(...))/h
=lim h->0 n*x^(n-1)+h*(...)
=n*x^(n-1)

[Anonymous No.5065698]

>>5065686
hmm the answer still didn't work...

[Anonymous No.5065715]

>>5065686

my god who can read that?

[Wolfram Book Guy No.5065842]

>>5065698
Assuming that you intended to refer to >>5065671 , look exactly what I gave you the derivate of and what the first step in the wolframalpha steps view said.

[Anonymous No.5066666]

Wait a sec, gonna post the solution with steps in a few minutes.

[Anonymous No.5066748]

>>5066666
NIGGER

[Anonymous No.5066756]

>>5066666
dem quints. also
<div class="math"> \frac{d}{dx} x^a = \frac{d}{dx} e^{a \log x } = e^{a \log x} \cdot a \frac{1}{x} = x^a \cdot a \frac{1}{x} = a \cdot x^{a-1 } </div> even for real a. or:
<div class="math"> \lim\limits_{x \rightarrow x_0 } \frac{ \frac{1}{ \sqrt{x} } - \frac{1}{ \sqrt{x_0} } }{x-x_0} = \lim\limits_{x \rightarrow x_0} \frac{ sqrt{x_0} -sqrt{x}}{ \sqrt{x} \sqrt{x_0} (x-x_0) } = \lim\limits_{x \rightarrow x_0 } \frac{1}{ \sqrt{x} \sqrt{x_0} } \cdot \frac{ \sqrt{x_0} - \sqrt{x} }{ (-1)(\sqrt{x_0} - \sqrt{x})(\sqrt{x_0} + \sqrt{x}) } = \lim\limits{x \rightarrow x_0} - \frac{1}{ \sqrt{x} \sqrt{x_0} } \cdot \frac{1}{ \sqrt{x_0} + \sqrt{x} } = - \frac{1}{x_0} \frac{1}{2 \sqrt{x_0} </div>

[Anonymous No.5066764]

>>5066666
dem quints. also
<div class="math"> \frac{d}{dx} x^a = \frac{d}{dx} e^{a \log x } = e^{a \log x} \cdot a \frac{1}{x} = x^a \cdot a \frac{1}{x} = a \cdot x^{a-1 } </div> even for real a
<div class="math"> \lim\limits_{x \rightarrow x_0 } \frac{ \frac{1}{ \sqrt{x} } - \frac{1}{ \sqrt{x_0} }}{x-x_0} = \lim\limits_{x \rightarrow x_0} \frac{ sqrt{x_0} -sqrt{x}}{ \sqrt{x} \sqrt{x_0} (x-x_0) } = \lim\limits_{x \rightarrow x_0 } \frac{1}{ \sqrt{x} \sqrt{x_0} } \cdot \frac{ \sqrt{x_0} - \sqrt{x} }{ (-1)(\sqrt{x_0} - \sqrt{x})(\sqrt{x_0} + \sqrt{x}) } = \lim\limits{x \rightarrow x_0} - \frac{1}{ \sqrt{x} \sqrt{x_0} } \cdot \frac{1}{ \sqrt{x_0} + \sqrt{x} } = - \frac{1}{x_0} \frac{1}{2 \sqrt{x_0}} </div>

[Anonymous No.5066857]

>>5065508
You fail at derivatives.

dy/dx=(-1/2)(7/x^(3/2))

dy/dx=-7/2[x^(3/2)]