What is the derivative of # x/(x^2+1)#?

1 Answer
Steve M
Dec 13, 2016

# d/dx (x)/(x^2+1) = ( 1-x^2 ) / (x^2+1)^2#

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #, or less formally, # " "(u/v)' = (v(du)-u(dv))/v^2 #

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with # y=(x)/(x^2+1) # Then

# { ("Let "u=x, => , (du)/dx=1), ("And "v=x^2+1, =>, (dv)/dx=2x ) :}#

# :. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #
# " "dy/dx = ( (x^2+1)(1) - (x)(2x) ) / (x^2+1)^2#
# " "dy/dx = ( x^2+1 - 2x^2 ) / (x^2+1)^2#
# " "dy/dx = ( 1-x^2 ) / (x^2+1)^2#

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