0
$\begingroup$

In my James Stewart Calculus book (Page 178) I'm completely confused by what this part means:

Notice that the limit is the value of the derivative of $f$ at 0, that is, $$\lim_{h \to 0} \frac{b^h - 1}{h} = f'(0)$$

Therefore we have shown that if the exponential function $f(x) = b^x$ is differentiable at 0, then it is differentiable everywhere and $$f'(x) = f'(0)*b^x$$

How does that limit equal f'(0) and how does that does tell us the exponential function is differentiable everywhere?

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

If $f(x) = b^x$ then $$f'(x) = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}=\lim_{h \to 0} b^x\frac{b^{h}-b^0}{h}=b^x\lim_{h \to 0} \frac{b^{h}-1}{h}=b^x f'(0).$$ Consequently, if $f'(0)$ exists then so does $f'(x) = b^x f'(0)$.

$\endgroup$
2
  • $\begingroup$ The part I dont understand is why f'(0) equals that limit? $\endgroup$
    – user98761
    Oct 14, 2017 at 20:58
  • 1
    $\begingroup$ @user32523850925 Put $x=0$ in $$f'(x) = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}.$$ $\endgroup$
    – Math Lover
    Oct 14, 2017 at 20:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .