Prove using the formal definition of a derivative that (f-g)'(x)= f'(x)-g'(x)?

Question

1045 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-04T09:59:41

See explanation.

According to the definition the derivative of $f (x)$ $f(x)$ is:

$f'(x)=lim_{h->0}(f(x+h)-f(x))/h$

If we apply the definition to $f(x)-g(x)$ we get:

$(f-g)'(x)=lim_{h->0}([f(x+h)-g(x+h)]-[f(x)-g(x)])/h$

$=lim_{h->0}(f(x+h)-g(x+h)-f(x)+g(x))/h$

$=lim_{h->0}(f(x+h)-f(x)-(g(x+h)-g(x)))/h$

$=lim_{h->0}(f(x+h)-f(x))/h-lim_{h->0}(g(x+h)-g(x))/h$

$=f'(x)-g'(x)$

QED