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How do you differentiate #y = (x) / (x+1)#?

Answer 1

Cameron Alexander

We can use the quotient rule, which states that for #y=f(x)/g(x)#, #(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2#

Following the formula:

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

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Answer 2

William Abdalla

To differentiate the function ( y = \frac{x}{x+1} ), we can use the quotient rule, which states that for a function ( \frac{u}{v} ), the derivative is given by ( \frac{u'v - uv'}{v^2} ), where ( u' ) represents the derivative of ( u ) with respect to ( x ) and ( v' ) represents the derivative of ( v ) with respect to ( x ). Applying this rule to ( y = \frac{x}{x+1} ), we have:

( u = x ) and ( v = x + 1 )

( u' = 1 ) (derivative of ( x )) and ( v' = 1 ) (derivative of ( x + 1 ))

Plugging these values into the quotient rule formula:

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

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

( y' = \frac{1}{(x+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.