Using the limit definition, how do you differentiate $f (x) = \frac{3}{x + 1}$ $f(x)= 3/(x+1)$ ?

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Answer 1 · 2015-11-28T01:53:10

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

The limit definition of the derivative:

$f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h$

We know that $f(x)=3/(x+1)$ , so $f(x+h)=3/(x+h+1)$ .

We can plug this in to find that:

$f'(x)=lim_(hrarr0)(3/(x+h+1)-3/(x+h))/h$

Find a common denominator.

$f'(x)=lim_(hrarr0)((3(x+1))/((x+h+1)(x+1))-(3(x+h+1))/((x+h+1)(x+1)))/h$

Simplify.

$f'(x)=lim_(hrarr0)(cancel(3x)cancel(+3)cancel(-3x)-3hcancel(-3))/(h(x+h+1)(x+1))$

$f'(x)=lim_(hrarr0)(-3cancel(h))/(cancel(h)(x+h+1)(x+1))$

$f'(x)=lim_(hrarr0)(-3)/((x+h+1)(x+1))$

Now, calculate the limit and plug in $0$ for $h$ .

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

Using the limit definition, how do you differentiate f(x)= 3/(x+1)f(x)=3x+1f(x)= 3/(x+1)?