Question

derivative: (x^3+1)^x (x^2+1)^(sinx)/x^(2cosx)

Answer 1

http://www.wolframalpha.com/input/?i=deritave+%28x%5E3%2B1%29%5Ex+%28x%5E2%2B1%29%5E%28sinx%29%2Fx%5E%282cosx%29

Answer 2

I really hope you typed that wrong.

Answer 3

nope, unfortunately that is the question :(

Answer 4

id up that last one with a negative exponent and product rule it into submission

Answer 5

\[D(fgh)=f'gh+fg'h+fgh'\]

Answer 6

thank you, i'll try that now

Answer 7

i think it makes life simpler in that you only have to focus on one part at a time ....

Answer 8

or try logarithmic differentiation...

Answer 9

yep

Answer 10

but that will be just as ugly

Answer 11

you still gotsta play the logs in the product; but no matter which route you go, just know that someone hates you enough to make you do this kind of problem :)

Answer 12

i've been working at it for 30 mins and it's so long and i got lost

Answer 13

and confused

Answer 14

Is this calc III?

Answer 15

cal 1

Answer 16

single variable derivatives are calc1

Answer 17

Yea but it is a hell of a problem, we never had anything like this is my calc 1 class.

Answer 18

georgia math, gotta love it :)

Answer 19

i know! I gave up halfway. it's so long and confusing. and frustrating.

Answer 20

you're from GA?

Answer 21

florida, but a had a foreman from georgia and i used to make fun of him for the way he would do his math. it was some backwards backwoods sorta stuff but he always got the right answer :)

Answer 22

haha

Answer 23

y = (x^3+1)^x

ln(y = (x^3+1)^x)

ln(y) = x ln(x^3+1)

y'/y = ln(x^3+1) + 3x^3/(x^3+1)

-------------------
y = (x^2+1)^(sinx)

ln(y=(x^2+1)^(sinx))

ln(y) = ln(x^2+1) sin(x)

y'/y = ln(x^2+1) cos(x) + 2x sin(x)/(x^2+1)

----------------------
y = x^(-2cosx)

ln(y = x^(-2cosx))

ln(y) = -2cos(x) ln(x)

y'/y = 2sin(x) ln(x) -2cos(x)/x

Answer 24

im not sure how far id try to simplify it after i put all the parts back in tho

Answer 25

i'll just skip this question. haha. thank you though