Question

Im having trouble with taking the derivative of this:
((x)/sqrt(x^2+1)

Answer 1

denominator is easy enough, it will be \(x^2+1\)
numerator is going to be a drag

Answer 2

If I used the product rule, it would be \[(x)(x^2+1)^(-1/2)\]
correct? Or would the x be a 1?

Answer 3

quotient rule is best product rule will make even more of a mess
it is really not that bad

Answer 4

I mean, I got \[\frac{ 1 }{(x^2+1)^3/2 }\]

Answer 5

numerator will be \(g'f-fg'\)with
\[f(x)=x,f'(x)=1,g(x)=\sqrt{x^2+1},g'(x)=\frac{x}{\sqrt{x^2+1}}\]

Answer 6

I don't know. I always feel as if Im doing them completely wrong haha.

Answer 7

plug and chug
you will get a compound fraction that you can take care of by multiplying top and bottom by \(\sqrt{x^2+1}\)

Answer 8

damn i made a typo above
numerator is \(gf'-fg'\)

Answer 9

write it out
numerator will be \(\sqrt{x^2+1}\times 1-x\times \frac{x}{\sqrt{x^2+1}}\)

Answer 10

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

Answer 11

denominator is \(x^2+1\)

Answer 12

so you get a nice ugly compound fraction
\[f'(x)=\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}\]

Answer 13

then multiply top and bottom by \(\sqrt{x^2+1}\) and i guess you get what you wrote above

Answer 14

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

Answer 15

Whoa! Yeah, I wasn't sure at all though. haha Thanks a ton! :D

Answer 16

Lol. I hope you realize how much that has helped me. lol
Im going through these like nothing. Thanks!