Author Topic: Some differentiation questions (Read 6298 times) Tweet Share

m@tty · « **Reply #30 on:** April 01, 2010, 12:48:17 pm »

No, wait. Here is a good one.... (sorry)

Differentiate
$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

No, not implicit, you already know that...

the.watchman · « **Reply #31 on:** April 01, 2010, 01:15:08 pm »

Yep, got that thanks!

I'm looking forward to using logarithmic diff!

Juddinator · « **Reply #32 on:** April 01, 2010, 02:04:06 pm »

Quote from: m@tty on April 01, 2010, 12:48:17 pm

No, wait. Here is a good one.... (sorry)

Differentiate
$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

No, not implicit, you already know that...

I'll post my answer up when I've completed my methods work for the day

the.watchman · « **Reply #33 on:** April 01, 2010, 02:08:35 pm »

Quote from: Juddinator on April 01, 2010, 02:04:06 pm

Quote from: m@tty on April 01, 2010, 12:48:17 pm
No, wait. Here is a good one.... (sorry)

Differentiate
$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

No, not implicit, you already know that...
I'll post my answer up when I've completed my methods work for the day

I won't post up mine (not really mine

)

m@tty · « **Reply #34 on:** April 01, 2010, 03:26:50 pm »

Quote from: the.watchman on April 01, 2010, 02:08:35 pm

Quote from: Juddinator on April 01, 2010, 02:04:06 pm
Quote from: m@tty on April 01, 2010, 12:48:17 pm
No, wait. Here is a good one.... (sorry)

Differentiate
$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

No, not implicit, you already know that...
I'll post my answer up when I've completed my methods work for the day

I won't post up mine (not really mine )

Lol, why not?

the.watchman · « **Reply #35 on:** April 01, 2010, 03:28:12 pm »

Quote from: m@tty on April 01, 2010, 03:26:50 pm

Quote from: the.watchman on April 01, 2010, 02:08:35 pm
Quote from: Juddinator on April 01, 2010, 02:04:06 pm
Quote from: m@tty on April 01, 2010, 12:48:17 pm
No, wait. Here is a good one.... (sorry)

Differentiate
$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

No, not implicit, you already know that...
I'll post my answer up when I've completed my methods work for the day

I won't post up mine (not really mine )

Lol, why not?

Because it's yours not mine

superflya · « **Reply #36 on:** April 01, 2010, 03:31:23 pm »

is there a quicker method to differentiate that?

the.watchman · « **Reply #37 on:** April 01, 2010, 03:34:18 pm »

Quote from: superflya on April 01, 2010, 03:31:23 pm

is there a quicker method to differentiate that?

m@tty showed me a logarithmic diff method of doing it

m@tty · « **Reply #38 on:** April 01, 2010, 03:41:28 pm »

It involves implicit diff, so it's Spesh stuff..

But the.watchman can elaborate on this method

the.watchman · « **Reply #39 on:** April 01, 2010, 03:44:49 pm »

Quote from: m@tty on April 01, 2010, 01:10:11 pm

Basically you take natural logs of both sides then differentiate both sides.

Using my example from the Methods board:

$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

Rather than differentiating this using the quotient rule then the product rule and chain rule, take $\log_e$ of both sides, simplify, then differentiate.

$\log_e{y}=\log_e{\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}}$

$\log_e{y}=\log_e{x^4\sqrt{x^2+1}}-\log_e{(2x+3)^6}$

$\log_e{y}=4\log_e{x}+\frac{1}{2}\log_e{x^2+1}-6\log_e{2x+3}$

Which is really easy to differentiate!

$\frac{d}{dx}(\log_e{y})=4\frac{d}{dx}(\log_e{x})+\frac{1}{2}\frac{d}{dx}(\log_e{x^2+1})-6\frac{d}{dx}(\log_e{2x+3})$

$\frac{dy}{dx}\cdot \frac{d}{dy}(\log_e{y})=\frac{4}{x}+\frac{2x}{2(x^2+1)}-\frac{12}{2x+3}$

$\frac{dy}{dx}\cdot \frac{1}{y}=\frac{4}{x}+\frac{x}{x^2+1}-\frac{12}{2x+3}$

$\frac{dy}{dx}=y\left(\frac{4}{x}+\frac{x}{x^2+1}-\frac{12}{2x+3}\right)$

Juddinator · « **Reply #40 on:** April 01, 2010, 03:55:11 pm »

Quote from: the.watchman on April 01, 2010, 03:44:49 pm

Quote from: m@tty on April 01, 2010, 01:10:11 pm
Basically you take natural logs of both sides then differentiate both sides.

Using my example from the Methods board:

$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

Rather than differentiating this using the quotient rule then the product rule and chain rule, take $\log_e$ of both sides, simplify, then differentiate.

$\log_e{y}=\log_e{\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}}$

$\log_e{y}=\log_e{x^4\sqrt{x^2+1}}-\log_e{(2x+3)^6}$

$\log_e{y}=4\log_e{x}+\frac{1}{2}\log_e{x^2+1}-6\log_e{2x+3}$

Which is really easy to differentiate!

$\frac{d}{dx}(\log_e{y})=4\frac{d}{dx}(\log_e{x})+\frac{1}{2}\frac{d}{dx}(\log_e{x^2+1})-6\frac{d}{dx}(\log_e{2x+3})$

$\frac{dy}{dx}\cdot \frac{d}{dy}(\log_e{y})=\frac{4}{x}+\frac{2x}{2(x^2+1)}-\frac{12}{2x+3}$

$\frac{dy}{dx}\cdot \frac{1}{y}=\frac{4}{x}+\frac{x}{x^2+1}-\frac{12}{2x+3}$

$\frac{dy}{dx}=y\left(\frac{4}{x}+\frac{x}{x^2+1}-\frac{12}{2x+3}\right)$

that's pro haha

and damn you guys and your ability to use Latex really fast! haha

superflya · « **Reply #41 on:** April 01, 2010, 04:01:40 pm »

yea i get that

m@tty · « **Reply #42 on:** April 01, 2010, 06:30:16 pm »

Quote from: Juddinator on April 01, 2010, 02:04:06 pm

Quote from: m@tty on April 01, 2010, 12:48:17 pm
No, wait. Here is a good one.... (sorry)

Differentiate
$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

No, not implicit, you already know that...
I'll post my answer up when I've completed my methods work for the day

You still going to differentiate it, Juddinator?

superflya · « **Reply #43 on:** April 01, 2010, 06:36:11 pm »

Quote from: m@tty on April 01, 2010, 06:30:16 pm

Quote from: Juddinator on April 01, 2010, 02:04:06 pm
Quote from: m@tty on April 01, 2010, 12:48:17 pm
No, wait. Here is a good one.... (sorry)

Differentiate
$y=\frac{x^4\sqrt{x^2+1}}{(2x+3)^6}$

No, not implicit, you already know that...
I'll post my answer up when I've completed my methods work for the day

You still going to differentiate it, Juddinator?

it'd take a good page of working using the quotient rule

m@tty · « **Reply #44 on:** April 01, 2010, 06:58:51 pm »

$\frac{(4x^3\cdot \sqrt{x^2+1}+\frac{x}{\sqrt{x^2+1}}\cdot x^4)(2x+3)^6-12(2x+3)^5\cdot x^4\cdot \sqrt{x^2+1}}{(2x+3)^{12}}$

Now simplify.

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Author Topic: Some differentiation questions (Read 6298 times) Tweet Share

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