Question

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

Answer 1

Remember that the limit definition of the derivative is defined as

`f'(x) = lim_(∆x->0) (f(x+∆x)-f(x))/(∆x)`

Therefore, since `f(x) = 2/x`, we write

`f'(x) = lim_(∆x->0) ((2/(x+∆x)) - (2/x))/(∆x)`

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

`a/b - c/d = (ad-bc)/(bd)`

So, we proceed by taking the limit as follows:

`f'(x) = lim_(∆x->0) (2x-2(x+∆x))/((x(x+∆x))/(∆x))`

`= 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:

`f(x) = 2x^(-1) -> f'(x) = -2x^(-2) = -(2)/(x^2)`

Answer 2

`-2/x^2`

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`