Question

find dy/dx if y = (x^2 + 1)^3x

Answer 1

write it as 3y=(x+(1/x))
now it looks easier i believe

Answer 2

The other way!!
\[ y = e^{\ln(x^2+1)*3x}\]

Answer 3

what's that before the 3x?

Answer 4

a plus?

Answer 5

so tiny

Answer 6

multiplication

Answer 7

asterisk * <--- stands for multiplication

Answer 8

im not sure what you did... you applied e to both sides?

Answer 9

No ... just \( a = e^{\ln a}\) ... just simple rule of exponents and lograthims

Answer 10

i just did chain rule so i guess that's not right

Answer 11

???

Answer 12

i got y' = 6x^2(x^2+1)^3x-1

Answer 13

http://www.wolframalpha.com/input/?i=%28differentiate+e^%283x+*+ln%28x^2%2B1%29%29%29+%3D+%28differentiate+%281%2Bx^2%29^%283x%29%29

Answer 14

you should be getting y'=(x^2-1)/(3x^2)

Answer 15

@experimentX I think your method is a lil bit complicated for a beginner in calculus

Answer 16

I think either you or i misread question
http://www.wolframalpha.com/input/?i=differentiate+%281%2Bx^2%29^%283x%29

Answer 17

its power aa
I misread the question my bad

Answer 18

im lost.

Answer 19

apply ln on both sides

Answer 20

it becomes lny=(3x)(ln(x^2+1))

Answer 21

now differentiate on both sides

Answer 22

why do i have to apply ln

Answer 23

it makes life easier

Answer 24

because exponent is 3x?

Answer 25

because there is an exponent

Answer 26

if this was explicit differentiation then i could just use chain rule, correct?

Answer 27

i got y' = 3xyln(x^2 + 1)

Answer 28

put back the values of y..

Answer 29

y' = 3x(x^2 + 1)^3x * ln(x^2 + 1)

Answer 30

Take log on both sides
\[ \ln y = 3x \ln(1+x^2)\]
Differentiate both sides
\[ \frac{1}{y} \frac{dy}{dx} = \frac{(6 x^2)}{(x^2+1)}+3 log(x^2+1)\]
\[ y' = y \left ( \frac{(6 x^2)}{(x^2+1)}+3 log(x^2+1) \right )\]
Put back the value of y,
\[ y' = (1 + x^2)^{3x} \left ( \frac{(6 x^2)}{(x^2+1)}+3 log(x^2+1) \right )\]
Answer!!!

Answer 31

oh i forgot to differentiate right side

Answer 32

log is the same as ln in this case?

Answer 33

yeah it's ln

Answer 34

ok! finally got it.