Question

How do you use the definition of a derivative to find the derivative of `f(x) = |x|`?

Answer 1

`f'(x) = { (-1, x<0), ("undefined",x=0), (1,x>0) :}`

Explanation:

By definition of the derivative `f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
So with `f(x) = |x|` we have;

`f'(x) = lim_(h rarr 0) ( |x+h| - |x| ) / h`
`:. f'(x) = lim_(h rarr 0) (( |x+h| - |x| )) / h (( |x+h| + |x| ))/(( |x+h| + |x| ))`
`:. f'(x) = lim_(h rarr 0) ( (|x+h|)^2 -(|x|)^2) / ( h( |x+h| + |x| ))`
`:. f'(x) = lim_(h rarr 0) ( (x+h)^2 -x^2) / ( h( |x+h| + |x| ))`
`:. f'(x) = lim_(h rarr 0) ( x^2+2hx+h^2 -x^2) / ( h( |x+h| + |x| ))`
`:. f'(x) = lim_(h rarr 0) ( 2hx+h^2 ) / ( h( |x+h| + |x| ))`
`:. f'(x) = lim_(h rarr 0) ( 2x+h ) / ( |x+h| + |x| )`
`:. f'(x) = ( 2x + 0) / ( |x+0| + |x| )`
`:. f'(x) = ( 2x ) / (2 |x| )`
`:. f'(x) = x / |x|`

Which if you think about it for a moment, we can write as;
`f'(x) = { (-1, x<0), ("undefined",x=0), (1,x>0) :}`

The above maths is actually a lot easier if you go straight to the definition of `|x|`, as in:

`|x| = { (-x, x<0), (0 , x=0), (x, x>0) :}`

And if we graph `y=|x|`, the result should be obvious, as is the reason for f'(x) being undefined at `x=0` (despite `y=|x|` being continuous when `x=0`)

graph{|x| [-10, 10, -5, 5]}

As, `{ ("differentiability" ,rArr, "continuity"), ("continuity", !rArr, "differentiability") :}`