Derivative simplification (a tangle!)

**rortiz81** · Jul 19th 2012, 06:06 PM

This thing is a monster:

Find $dy/dx$ for $y=\frac{sin^3(5x)\left(4x-x^2\right)^\frac{1}{3}}{1-tan(x)}$

Is there some kind of easy way to differentiate this thing? Am I missing something?

So far I have $y=(st)\left(u\right)^\left(-1\right)$ , where $s=sin^3(5x), t=\left(4x-x^2\right)^\frac{1}{3}, u=1-tan(x)$

Product rule, chain rule, blah blah ... I get: $y'=(st)\left(\left(u\right)^\left(-1\right)\right)^'$ $+\left(st\right)^'$ $\left(u\right)^\left(-1\right)$ $=-(st)\left(u\right)^\left(-2\right)$ $\left(u\right)^'\right)\right)+\left(st'+ts'\right )\left(u\right)^\left(-1\right)$ Factoring: $y'=\left(u\right)^\left(-1\right)$ $\right)\left(st'+ts'-stu'\left(u\right)^\left(-1\right)\right)$

My individual derivatives are $s'=15sin^2(5x)cos(5x), t'=\frac{4-2x}{3\left(4-x^2\right)^\frac{2}{3}}, u'=-sec^2(x)$

SOOOOO putting it all together I get: $y'=-\frac{\left(\frac{sin^3(5x)\left(4-2x\right)}{3\left(4-x^2\right)^\frac{2}{3}}+15sin^2(5x)cos(5x)+\frac{s in^3(5x)\left(4x-x^2\right)^\frac{1}{3}sec^2(x)}{1-tan(x)}\right)}{1-tan(x)}$

And I can factor futher:

$y'=-\frac{sin^2(5x)\left(4x-x^2\right)^\frac{1}{3}\left(\frac{sin(5x)\left(4-2x\right)}{12-3x^2}+15cos(5x)+\frac{sin(5x)sec^2(x)}{1-tan(x)}\right)}{1-tan(x)}$

Does this seem right? Please help I have very little time to submit this and my teacher is a ... ah ... hard-case. She'll mark me down for skipping steps, despite the right answer. Can this be futher simplified? Was there an easier way to go about this?

Thanks for any help you guys can provide!

**tom@ballooncalculus** · Jul 20th 2012, 02:13 AM

Comparing here...

sin^3(5x) (4x - x^2)^(1/3) (1 - tan x)^(-1) - Wolfram|Alpha

... you'll see your only final error is your overall negative sign (where from?). Although earlier (in the SOOOOO line) you're missing the polynomial part of the middle term.

The best chance of minimising tangles in this case is probably ...

Product rule - Wikipedia, the free encyclopedia

... from which you get this overview...

... the bottom line of which is pretty much Wolfram's final version.

(Straight lines differentiate downwards with respect to x.)

If there's a better way I don't see it. Edit: D'oh! Yes of course, thanks Skeeter! Although, funnily enough, the log method doesn't seem any simpler in this case. Though it often is, obviously.

Larger

I probably won't zoom in on the chain rules on this occasion - it might just re-aggravate the tangle. Not promising, though!

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**skeeter** · Jul 20th 2012, 04:22 AM

primo candidate for logarithmic differentiation ...

$y = \frac{\sin^3(5x) \sqrt[3]{4x-x^2}}{1-\tan{x}}$

$\ln{y} = 3\ln[\sin(5x)] + \frac{1}{3}\ln(4x-x^2) - \ln(1 - \tan{x})$

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{15\cos(5x)}{\sin(5x)} + \frac{4-2x}{3(4x-x^2)} + \frac{\sec^2{x}}{1-\tan{x}}$

$\frac{dy}{dx} = \frac{\sin^3(5x) \sqrt[3]{4x-x^2}}{1-\tan{x}} \left[15\cot(5x) + \frac{2(2-x)}{3x(4-x)} + \frac{1+\tan^2{x}}{1-\tan{x}}\right]$

**rortiz81** · Jul 20th 2012, 05:23 PM

Thanks guys! I re-worked the problem and got:

$\frac{dy}{dx}=sin^2(5x)(4x-x^2)^\frac{1}{3}\left(\frac{15cos(5x)}{1-tan x}+\frac{sin(5x)(4-2x)}{3(1-tan x)(4x-x^2)}+\frac{sin(5x)}{sin^2 x}\right)$

Which isn't that far off; I'm a little worried about that $cos(5x)$ in place of skeeter's $cot(5x)$ in the first numerator but the rest seems ok. My weakness (one of em, anyway) is trigonometric proofs, so reworking 'em gives me headaches. That's what I submitted so hopefully I won't lose too many points for it.

Thank skeeter, thanks tom!!

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