Question

Derivative of f(x) = 3/x using limit definition of derivative?

Answer 1

`f'(x)=-3/x^2`

Explanation:

The definition of a derivative

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

We want to differentiate `f(x)=3/x`, therefore we seek

`f'(x)=lim_(h->0)(3/(x+h)-3/x)/h`

Thus

`f'(x)=3lim_(h->0)(1/(x+h)-1/x)/h`

`=3lim_(h->0)(1/(h(x+h))-1/(hx))`

`=3lim_(h->0)(x-(x+h))/(h(x+h)x)`

`=3lim_(h->0)(-h)/(h(x+h)x)`

`=3lim_(h->0)(-1)/((x+h)x)`

`=3lim_(h->0)(-1)/(x^2+hx)`

`=-3/x^2`

Answer 2

Given: `f(x) = 3/x` then `f(x+h) = 3/(x+h)`

The limit definition of the derivative is:

`f'(x) = lim_(h to 0) (f(x+h)-f(x))/h`

Substitute `f(x) = 3/x` and `f(x+h) = 3/(x+h)`:

`f'(x) = lim_(h to 0) (3/(x+h)-3/x)/h`

Multiply by 1 in the form of `(x(x+h))/(x(x+h))`:

`f'(x) = lim_(h to 0) (x(x+h))/(x(x+h))(3/(x+h)-3/x)/h`

Perform the multiplication:

`f'(x) = lim_(h to 0) ((3x(x+h))/(x+h)-(3x(x+h))/x)/(h(x(x+h)))`

Please observe how this simplifies the numerator by cancellation:

`f'(x) = lim_(h to 0) ((3xcancel((x+h)))/cancel((x+h))-(3cancel(x)(x+h))/cancel(x))/(h(x(x+h)))`

The limit with the cancelled factors removed:

`f'(x) = lim_(h to 0) (3x-3(x+h))/(h(x(x+h)))`

Distribute the -3 in the numerator:

`f'(x) = lim_(h to 0) (3x-3x-3h)/(h(x(x+h)))`

Combine like terms:

`f'(x) = lim_(h to 0) (-3h)/(h(x(x+h)))`

`h/h` becomes 1:

`f'(x) = lim_(h to 0) (-3)/(x(x+h))`

Now, we may let `h to 0`:

`f'(x) = -3/(x(x))`

Simplify:

`f'(x) = -3/x^2`