Question

18. Let f be a function with domain the set of all real numbers and having the following

Answer 1

The derivative of the given function is f'(x) = k f(x) where $k= \lim_{h \to 0} \frac{f(h)-1}{h}$.

What is the derivative of a function?

Let f be a function defined on a neighborhood of a real number a. Then f is said to be differentiable or derivable at 'a' if $\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ exists finitely. The limit is called the derivative or differential coefficient of f at 'a'. It is denoted by f'(a).

If f is differentiable at 'a', then

$f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}$

Calculation:

The given properties are:

(i) f(x + y) = f(x)f(y) for all real numbers x and y.

(ii) $\lim_{h \to 0} \frac{f(h)-1}{h}$ = k; where k is a nonzero real number.

Then, the derivative of the function f(x) is,

f'(x) = $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

From property (i), f(x + h) = f(x)f(h)

On substituting,

f'(x) = $\lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h}$

      = $\lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}$

From property (ii), $\lim_{h \to 0} \frac{f(h)-1}{h}$ = k;

f'(x) = $\lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}$

      = f(x). $\lim_{h \to 0} \frac{f(h)-1}{h}$

      = f(x). k

      = kf(x)

Therefore, f'(x) = k f(x); where f'(x) exists for all real numbers of x.

Learn more about the derivative of a function here:

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