Question

Differentiate f(x)=(x=1)/(x^2+1)^1/2

i keep getting [(x^2+1)^1/2 - x(x+1)] / (x^2+1)^1/2
but wolramalpha says it is something else can you tell me what im doing wrong?

Answer 1

i think they are multiply the sqrt root with a common denominator

Answer 2

So is the function\[f(x) = \frac{ x -1}{ \sqrt{x^{2}+1} }\]

Answer 3

yes !

Answer 4

you'll need to apply quotient rule

Answer 5

i did and i got
\[\sqrt{x^2+1}-\frac{ x(x+1) }{ \sqrt{x^2+1} }\]

Answer 6

and all that divided by (x^2+1)

Answer 7

I think the denominator must not be a radical.

Answer 8

so what do i do multiply by a radical?

Answer 9

\[\frac{ d }{ dx }(\frac{ x-1 }{ \sqrt{x^{2}+1} }) = \frac{\sqrt{x^{2}+1}d(x-1)-(x-1)d(\sqrt{x^{2}+1}) }{ (\sqrt{x^{2}+1})^{2} }\]

Answer 10

where did the (x-1) come from?

Answer 11

The quotient rule tells that
\[d(\frac{ u }{ v }) = \frac{ vdu - udv }{ v^{2} }\]

Answer 12

but the original problem is x+1 not x-1

Answer 13

you said yes when i ask you if what i wrote is the right given. :(
well, anyway, just change x-1 to x+1

Answer 14

omg im sorry. and yeah i get it the derivative thing. what i do not understand is how can i simplify it? how do i get rid of the radical on the denominator?

Answer 15

you must multiply it by itself. that's the quotient rule

Answer 16

what i did is muliply the sqrt(x^2+1) by itself so that it would have a common denominator with what im subtracting it with and got
\[\frac{ x^2+1-x^2-x }{ \sqrt{x^2+1} } / x^2+1\]

Answer 17

the a is supposed to be divided by

Answer 18

\[\int\limits \frac{x+1}{\left(x^2+1\right)^{3/2}} \, dx=\frac{x-1}{\sqrt{x^2+1}} \]