Let f and g be differentiable at a point x0εDom(f)nDom(g) show that
(1) (f+g)'(x0)=f'(x0)+g'(x0)
(2) (fg)'(x0)=f'(x0)g(x0)+f(x0)g'(x0)
(3) (cf)'(x0)=cf'(x0), if cεR
(1) (f+g)'(x0)=f'(x0)+g'(x0)
(2) (fg)'(x0)=f'(x0)g(x0)+f(x0)g'(x0)
(3) (cf)'(x0)=cf'(x0), if cεR
SOLUTION
(1) since f(x) is differentiable at x0, we know from the derivative formula that
Here we can see that x→x0, we can expand the formula since the functions are differentiable at x0, by simply asumming x0→0
Now substitute (f+g) '(x) into the derivative formula and we will have
Simplify the fraction by breaking the numerator into two parts, and recall from the rules of Limits that the limits of a sum Is the sum of the limits. Hence
By definition we have
And hence we have
Hence our argument is proved.
(2) (fg)'(x0)=f '(x0)g(x0)+g'(x0)f(x0)
This is the product formula of differentiation, we are going to use the same concept of differentiability to proof our argument but in a composite form.
Substituting f and g into the formula above, we get
In order to keep our proof logical, all we need to do is to expand the numerator in the sense that we just add and subtract f(x+x0)g(x) and then we can factor out.
By using the properties of Limits, we get
And we know from the formula of derivative that
And hence we get
(fg)'(x0)=f'(x0)g(x0)+f(x0)g'(x0)
(3) (cf)'(x0)=cf'(x0), if cεR
Here we can see that c is a real number, and f'(x0) is a derivative. All we need to do here is simply to show that our Proof is logical.
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