Question

Find the derivative using rules.

(x^2+3)/(x-1)^3+(x+1)^3

Answer 1

Is the question
(i) \(\frac{(x^2+3)}{(x-1)^3+(x+1)^3}\)
or
(ii) \(\frac{(x^2+3)}{(x-1)^3}+(x+1)^3\)?

Answer 2

First one

Answer 3

Okay.
Do you know quotient rule and chain rule?

Answer 4

quotient rule yes, we have not learned the chain rule yet...

Answer 5

Do you know how to differentiate \((x-1)^3\)?

Answer 6

just use power rule lol

Answer 7

\[3(x-1)^{2}\]?

Answer 8

^

Answer 9

Yup.
What's (x-1)^3 + (x+1)^3?

Answer 10

3(x−1)^2 + 3(x+1)^2

Answer 11

And (x^2+3) ?

Answer 12

2x

Answer 13

Ok. Then you're ready to apply quotient rule!

\[\frac{d}{dx}(\frac{u}{v}) = \frac{vu'-uv'}{v^2}\]

Let say, in your question, u = x^2+3, what is v?

Answer 14

.3(x−1)^2 + 3(x+1)^2? o_o

Answer 15

lol

Answer 16

No o_o

Answer 17

Compare \(\huge\frac{u}{v}\) and \(\huge \frac{(x^2+3)}{(x-1)^3+(x+1)^3}\)

If u = x^2+3
what is v?

Answer 18

The denominator...(x-1)^3 + (x+1)^3

Answer 19

Yes!!!!!
Now, u' is the derivative of u(with respect to x); v' is the derivative of v (with respect to x).

u =x^2+3
v = (x-1)^3 + (x+1)^3

u' = ???
v' = ???

Answer 20

u = 2x
v = 6(x^2+1)?

Answer 21

u' v'. Correction :)

Answer 22

Noooooo. Check v' again!!

Answer 23

Finding v' is the same as finding the derivative of (x-1)^3 + (x+1)^3

Answer 24

I'm lost then. Why isn't it 3(x−1)^2 + 3(x+1)^2

Answer 25

It is.
u =x^2+3
u' = 2x
v = (x-1)^3 + (x+1)^3
v' = 3(x−1)^2 + 3(x+1)^2

Now, plug everything into
\[\frac{d}{dx}(\frac{u}{v}) = \frac{vu'-uv'}{v^2}\]
Then you can get your answer.

Answer 26

Great, thanks!

Answer 27

Welcome~