Question

Help with derivatives

Answer 1

\[y=\sqrt[4]{2x ^{3}+ 2x ^{2}-x}\]

Answer 2

write it as \(u^{\frac{1}{4}}\) so that \(du=\frac{d}{dx}u^{\frac{1}{4}}du\)

Answer 3

why the u?

Answer 4

I don't understand what to do

Answer 5

We just need a variable to stand for the stuff in the middle. Then we can do something like
\(\large \frac{d}{dx}=\sqrt[4]{2x^3+2x^2-x}=\frac{1}{4}(2x^3+2x^2-x)^{\frac{1}{4}-1}*\frac{d}{dx}2x^3+2x^2-x\)

Answer 6

chain rule

Answer 7

ahhh ok

Answer 8

so \[\frac{ 1 }{ 4 } (2x ^{3}+2x ^{2}- x)^{\frac{ 3 }{ 4 }} * 6x ^{2} + 4x -1\]

Answer 9

sorry it should be -3/4

Answer 10

@wio that looks right, right?

Answer 11

Yes, it looks right.

Answer 12

But you need parenthesis.

Answer 13

and we can simplify it as a single fraction \(\large \frac{6x^2+4x-1}{4(2x^3+2x^2-x)^\frac{3}{4}}\)

Answer 14

Technically you should have said: \[
dy =\frac{d}{du} \sqrt[4]{u}\;du
\]

Answer 15

oops

Answer 16

ok so what do I write?

Answer 17

Well, you already got it, just make sure that you put things in parenthesis to maintain order of operations.

Answer 18

what does that mean?

Answer 19

wrap parentheses around the \([\huge \frac{ 1 }{ 4 } (2x ^{3}+2x ^{2}- x)^{\frac{- 3 }{ 4 }}] * 6x ^{2} + 4x -1\)

Answer 20

ohhhhhh ok that's fine, I've that written on my page

Answer 21

I'm just not used to typing my equations electronically :/

Answer 22

well, it looks prettier if you have it as a single fraction

Answer 23

totes xox

Answer 24

Can I just leave it as that or is more to be done?

Answer 25

\(\large [\frac{ 1 }{ 4 } (2x ^{3}+2x ^{2}- x)^{\frac{- 3 }{ 4 }}] * 6x ^{2} + 4x -1= \frac{6x^2+4x-1}{4(2x^3+2x^2-x)^\frac{3}{4}}\)

Answer 26

is there anything that has to be done to the fraction?

Answer 27

?

Answer 28

not as far as I can tell

Answer 29

Ok well thank you :)

Answer 30

np :)