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How do you differentiate #f(x)=(2x)/(x^2+1)#?

Answer 1

Christopher Agresta

#d f(x)=(2 *d x)/(x^2+1)#

#d f(x)=(2x)/(x^2+1)*d x#

#d f(x)=(2(x^2+1)+2x*2x)/(x^2+1)^2 *d x#

#d f(x)=(2(x^2+1)+4x^2)/(x^2+1)^2 * d x#

#d f(x)=(2x^2+2+4x^2)/(x^2 +1)^2 *d x#

#d f(x)=(6x^2+2)/(x^2+1)^2 *d x#

#d f(x)=(2cancel((x^2+1)))/cancel((x^2+1))^2* d x#

#d f(x)=(2 *d x)/(x^2+1)#

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Answer 2

Emma Aldridge

To differentiate ( f(x) = \frac{2x}{x^2 + 1} ), you can use the quotient rule. The quotient rule states that if you have a function ( \frac{u(x)}{v(x)} ), then its derivative is given by ( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ). Applying this to ( f(x) ), where ( u(x) = 2x ) and ( v(x) = x^2 + 1 ), you get:

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

[ f'(x) = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} ]

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