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How do you find the derivative of #(x)=(x)/(x^2+1)#?

Answer 1

Emily Ammerman

#f=x,g=x^2+1,f'=1,g'=2x->f'(x)=((gf'-fg')/g^2) =(x^2+1-2x^2)/(x^2+1)^2 = (1-x^2)/(x^2+1)^2#

Separate f and g and find their derivatives then plug in to the quotient rule and simplify

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Answer 2

Aurora Alvarado

To find the derivative of ( f(x) = \frac{x}{x^2 + 1} ), you can use the quotient rule. The quotient rule states that if ( f(x) = \frac{g(x)}{h(x)} ), then ( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} ). Applying this rule:

( g(x) = x ) and ( h(x) = x^2 + 1 ).

( g'(x) = 1 ) (derivative of (x) is (1)) and ( h'(x) = 2x ) (derivative of (x^2 + 1) is (2x)).

Now plug these into the quotient rule:

( f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} )

( = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} )

( = \frac{-x^2 + 1}{(x^2 + 1)^2} )

So, the derivative of ( f(x) = \frac{x}{x^2 + 1} ) is ( f'(x) = \frac{-x^2 + 1}{(x^2 + 1)^2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.