Question

Please help. 4 derivative problems.

Answer 1

Here's the first one: \[\frac{ \sqrt[3]{x} }{ 1+x }\]

Answer 2

use the quotient rule

\[\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\]

Answer 3

Identify \(u\) and \(v\),

then find \(u'\), \(v'\), and \(v^2\)

Answer 4

Here's what I got so far. Could you kindly check. \[\frac{ (1+x)\frac{ 1 }{ 3 } (x)^{\frac{ -2 }{ 3 }} - (\sqrt[3]{x}) }{ (1+x)^{2} }\]

Answer 5

yes, that is correct, now simplify

Answer 6

Ok, thanks! Please help me simplify though. Haha. I'm not sure if I'd be right.

Answer 7

simplify this bit first \[(1+x)\frac{ 1 }{ 3 } (x)^{\frac{ -2 }{ 3 }}\]

Answer 8

\[\frac{ x ^{\frac{ -2 }{ 3 }} + x ^{\frac{ 1 }{ 3 }} }{ 3 }\]

Answer 9

How's that?

Answer 10

good,

now combine that other term in the numerator

Answer 11

Oh, yeah. So it's x^-1/3 ?

Answer 12

um \(-\sqrt[3]x\) is equal to \(-x^{1/3}\)

Answer 13

\[\frac{ (1+x)\frac{ 1 }{ 3 } (x)^{\frac{ -2 }{ 3 }} - (\sqrt[3]{x}) }{ (1+x)^{2} }\]
\[\frac{ \frac{ x ^{-2/3} + x ^{1/3} }{ 3 } -x^{1/3} }{ (1+x)^{2} }\]

Answer 14

Oh wait, back up. Haha. My last answer was I just combined the numerator.

Answer 15

With the Xs, with the exponent, the one with the 3 denominator.

Answer 16

Uncle, what about substitution ?

Answer 17

Guys, I'll close this now. I'll just stick to my answer. Hope it's right. lol. Thanks!

Answer 18

http://www.wolframalpha.com/input/?i=differentiate+cube+root+of+x+%2F+%281%2Bx%29+

This is what wolfram gives as the answer.

Answer 19

ah so we are close to the answer, we just to finish the simplification (rearrangement )

Answer 20

\[\frac{ \frac{ x ^{-2/3} + x ^{1/3} }{ 3 } -x^{1/3} }{ (1+x)^{2} }\]
\[\frac{x^{-2/3} + x ^{1/3} -3x^{1/3} }{ 3(1+x)^{2} }\]\[...\]

Answer 21

If we use Substitution : cube root of x = u

diff. both sides ... \(dx = \cfrac{du}{\cfrac{1}{3} x^{-2/3} } \)

\(\cfrac{d}{dx} \left( \cfrac{u}{1+u^3} \right) \times \cfrac{1}{3}x^{-2/3} \)

\(\cfrac{(1+x) - x^{1/3} (3 \times x^{2/3} ) }{(x+1)^2} \times \cfrac{1}{3} x^{-2/3} \)

\(\cfrac{1-2x}{3(x+1)^2 \times x^{2/3} } \)

Answer 22

\[\frac{x^{-2/3} + x ^{1/3} -3x^{1/3} }{ 3(1+x)^{2} }\]
\[\frac{x^{-2/3} -2x ^{1/3}}{ 3(1+x)^{2} }\]
\[\frac{(1 -2x)x^{-2/3}}{ 3(1+x)^{2} }\]
\[\frac{(1 -2x)}{ 3(1+x)^{2}x^{2/3} }\]

the answers all match
yayayay\(\checkmark\checkmark\checkmark\)

Answer 23

Well Done Mr. Rhaukus :P