Differentiate (x^2-1)/(x^2+1)?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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

Let f(x) = u/v such that:

u = (x^2 -1)  ==> u' = 2x

v= x^2 + 1  ==>   v' = 2x

==> f'(x)= (u'v-uv')/v^2

              = (2x(x^2 + 1) - (x^2 -1)2x ]/(x^2 +1)^2

              = (2x^3 + 2x - 2x^3 + 2x)/(x^2+1)^2

              = 4x/(x^2 + 1)

==> f'(x) = 4x/(^2 + 1)^2

 

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll differentiate applying quotient rule.

y = (x^2-1)/(x^2+1)

y = u(x)/v(x)

We'll differentiate:

y' = [u'(x)*v(x)-u(x)*v'(x)]/v(x)^2

v(x) = (x^2+1)

 v'(x) = 2x

u(x) = x^2-1 .

u'(x) = (x^2-1)'

u'(x) = 2x

y' = [2x*(x^2+1)- 2x*(x^2-1)]/(x^2+1)^2

We'll remove the brackets:

y' = (2x^3 + 2x - 2x^3 + 2x)/(x^2+1)^2

We'll eliminate like terms:

y' = (4x)/(x^2+1)^2

The result of differentiation is: y' = (4x)/(x^2+1)^2

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To differentiate  y = (x^2-1)/(x^2+1)

We use {u(x)/v(x)}' = {u'(x)v(x)-u(x)v(x)}/(v(x))^2 to fifferentiate the given expression.

u(x) = (x^2-1). u'(x) = (x^2-1)' = 2x

v(x) = (x^2+1). So (x^2+1'  = 2x.

Therefore {x^2-1)/(x^2+1)}' = {(x^2-1)(x^2+1) -(x^2-1)(x^2+1)'}/(x^2+1)^2

{(x^2-1)/(x^2+1) }'= {2x(x^2+1)-(x^2-1)*2x}/(x^2+1)

{(x^2-1)/(x^2+1)}' = 4x/(x^2+1)^2

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