/sci/ - Science & Math » Thread #14059243

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help :)

bexy Wed 05 Jan 05:49:32 2022 No.14059243 View Reply Original Report

Quoted By: >>14059249 >>14059250 >>14059532 >>14060200 >>14061172

i need help for my problem

Anonymous

Anonymous Wed 05 Jan 2022 05:52:22 No.14059249 Report

Quoted By: >>14059263

>>14059243
f(x)=(3x^2+1)/x=3x+1/x
f'(x)=3+lnx

Anonymous

Anonymous Wed 05 Jan 2022 05:52:23 No.14059250 Report

Quoted By:

>>14059243
Quotient rule

Anonymous

Anonymous Wed 05 Jan 2022 05:52:44 No.14059253 Report

Quoted By: >>14059280 >>14059287

f(x) = 3x^2/x + 1/x
f(x) = 3x + 1/x
because the derivative of a sum is the sum of the derivatives
f'(x) = 3 + ln(x)

Anonymous

Anonymous Wed 05 Jan 2022 05:55:08 No.14059263 Report

Quoted By: >>14059283

>>14059249
What?

Anonymous

Anonymous Wed 05 Jan 2022 05:59:05 No.14059280 Report

Quoted By: >>14059283

>>14059253
That is wrong, by the quotient rule, the derivative is 3-(1/x^2)

Anonymous

Anonymous Wed 05 Jan 2022 06:00:13 No.14059283 Report

Quoted By:

>>14059263
My mistake, what this guy said >>14059280

Anonymous

Anonymous Wed 05 Jan 2022 06:01:11 No.14059287 Report

Quoted By:

>>14059253
ln(x) is the integral of 1/x not the derivative.

Anonymous

Anonymous Wed 05 Jan 2022 06:54:05 No.14059410 Report

Quoted By:

I'm here, anon. What is your problem?

Anonymous

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Anonymous Wed 05 Jan 2022 08:01:46 No.14059532 Report

Quoted By: >>14060509 >>14060592

>>14059243
>problem requires quotient rule
>use chain rule and product rule instead
Who else /devilish/?

Anonymous

Anonymous Wed 05 Jan 2022 12:05:26 No.14060083 Report

Quoted By:

f(x) = 3 - 1/x2

Anonymous

Anonymous Wed 05 Jan 2022 12:07:22 No.14060090 Report

Quoted By:

The derivative of 1/x is -1/x^2

Anonymous

Anonymous Wed 05 Jan 2022 12:38:07 No.14060194 Report

Quoted By: >>14060209

y = (3x2 + 1) / x

adding nudge to function
y + dy = (3(x + dx)2 + 1) / (x + dx)

expanding binomial
y + dy = (3x2 + 6xdx +dx2 + 1) / (x + dx)

subtracting y from right side
dy = (3x2 + 6xdx +dx2 + 1) / (x + dx) - (3x2 + 1) / x

common denominator
dy = x(3x2 + 6xdx + dx2 + 1) / (x (x + dx)) - (3x2 + 1)(x + dx) / (x (x + dx))

series of simplifications
dy = (3x2dx + xdx2 - dx) / (x (x + dx))

derivative function
dy = dx(3x2 + xdx - 1) / (x2 + xdx)
dy/dx = (3x2 + xdx - 1) / (x2 + xdx)

as dx approaches 0,
dy/dx = (3x2 - 1) / x2
= 3 - x-2

Anonymous

Anonymous Wed 05 Jan 2022 12:39:07 No.14060200 Report

Quoted By:

>>14059243
$\ln \left (f(x) \right ) = \ln \left ( 3x^2 + 1 \right ) - \ln (x)$
$\text{let} ~ u = 3x^2 + 1 \implies \frac{d}{dx}\ln\left(u(x) \right) = \frac{1}{u(x)}\frac{du}{dx} = \frac{6x}{3x^2+1}$
So: $\frac{1}{f(x)}\frac{df}{dx} = \frac{6x}{3x^2+1}-\frac{1}{x}$
$\therefore \frac{df}{dx} =6 - \frac{3x^2 +1}{x^2} = 3 - \frac{1}{x^2}$ Simple!

Anonymous

Anonymous Wed 05 Jan 2022 12:40:39 No.14060209 Report

Quoted By:

>>14060194
shit formatting
y = (3x^2 + 1) / x
y + dy = (3(x + dx)^2 + 1) / (x + dx)
y + dy = (3x^2 + 6xdx +dx^2 + 1) / (x + dx)
dy = (3x^2 + 6xdx +dx^2 + 1) / (x + dx) - (3x^2 + 1) / x
dy = x(3x^2 + 6xdx + dx^2 + 1) / (x (x + dx)) - (3x^2 + 1)(x + dx) / (x (x + dx))
dy = (3x^2dx + xdx^2 - dx) / (x (x + dx))
dy = dx(3x^2 + xdx - 1) / (x^2 + xdx)
dy/dx = (3x^2 + xdx - 1) / (x^2 + xdx)

as dx approaches 0,
dy/dx = (3x^2 - 1) / x^2
= 3 - x^-2

Anonymous

Anonymous Wed 05 Jan 2022 14:06:23 No.14060509 Report

Quoted By:

>>14059532
Kek I never seem to remember the quotient rule so I just do this

Anonymous

Anonymous Wed 05 Jan 2022 14:25:49 No.14060592 Report

Quoted By:

>>14059532
>problem requires product rule
>use quotient rule and reciprocal rule instead
ONE
NOTHING'S WRONG WITH ME

Anonymous

Anonymous Wed 05 Jan 2022 16:45:45 No.14061172 Report

Quoted By: >>14061336

>>14059243
I was told this was the smart board but I come here to see if I could solve a math problem I would've done in first year (6+ years ago, I finished my le stem degree and never used it) and there are multiple people messing up basic derivatives.

Anonymous

Anonymous Wed 05 Jan 2022 16:48:42 No.14061184 Report

Quoted By:

>several people ITT confusing derivatives and intergrals
Brianlet board

Anonymous

Anonymous Wed 05 Jan 2022 17:33:14 No.14061336 Report

Quoted By:

>>14061172
yeah well it's been 15 years since i last did calculus and i forgot that ln x was the integral and not the derivative of 1/x
not like you need any of this for crypto trading anyway so who cares

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