# How do you use the limit definition of the derivative to find the derivative of #f(x)=2/x#?

##### 2 Answers

Remember that the limit definition of the derivative is defined as

Therefore, since

In the next step, we can use some algebraic properties such as

So, we proceed by taking the limit as follows:

#= lim_(∆x->0) (cancel(2x-2x)-2∆x)/((x^2+x∆x)/(∆x))#

#= lim_(∆x->0) ((-2cancel(∆x))/((x^2+x∆x)) * 1/(cancel(∆x)))/cancel(((∆x)/(1))*(1/(∆x)))#

#= lim_(∆x->0) (-2)/(x^2+x∆x)#

#= (-2)/(x^2+x(0))#

#=-2/(x^2)#

Checking our answer by using the power rule for derivatives:

#### Explanation:

To find the derivative of f(x) from 'first principles'.

#color(red)(|bar(ul(color(white)(a/a)color(black)(f'(x)=lim_(hto0)(f(x+h)-f(x))/h)color(white)(a/a)|))#

#f(x+h)=2/(x+h)" and " f(x)=2/x#

#rArrf'(x)=(2/(x+h)-2/x)/h# The aim is to eliminate the h from the denominator, as this would be undefined when h is zero.

Begin by simplifying the numerator, as a single fraction.

#rArrf'(x)=lim_(hto0)((2x)/(x(x+h))-(2(x+h))/(x(x+h)))/h#

#=lim_(hto0)(2x-2x-2h)/(hx(x+h)#

#=lim_(hto0)(-2h)/(hx(x+h))=lim_(hto0)(-2cancel(h))/(cancel(h)x(x+h))# The h is now eliminated from the denominator

#rArrf'(x)=-2/x^2#