Question

How can I describe the x-values at which f is differentiable at $f(x) = \dfrac{2}{{(x - 3)}}$? What is differentiable anyway?

Answer 1

Here we use the definition of differentiable function and find the derivative of the given function. Find all the values of x for which the function is differentiable i.e. for which the derivative exists.

• A function $f$is differentiable if the derivative exists for every value $a$in the domain.
  $f'(a) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$exits.
  General formula of differentiation is $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$

Complete step by step solution:
We have the function $f(x) = \dfrac{2}{{(x - 3)}}$ … (1)
We differentiate the function with respect to x
$\Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{d}{{dx}}\left( {\dfrac{2}{{(x - 3)}}} \right)$
$\Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{d}{{dx}}\left( {2{{(x - 3)}^{ - 1}}} \right)$
Use the general formula of differentiation $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$\Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = 2\left( { - 1{{(x - 3)}^{ - 1 - 1}}} \right)$
Add the terms in power
$\Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = - 2\left( {{{(x - 3)}^{ - 2}}} \right)$
$\Rightarrow \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{{ - 2}}{{{{(x - 3)}^2}}}$ … (2)
So, the derivative of the function exists.
We know the derivative obtained in equation (2) will not exist if the denominator is 0
$\Rightarrow {(x - 3)^2} = 0$
Then if few put the value of x as 3, then the denominator becomes 0.
Therefore, $f(x) = \dfrac{2}{{(x - 3)}}$ is differentiable at all values of x except at $x = 3$ .
A function is said to be differentiable if its derivative exists at all points of its domain. So, differentiable function obtains some constant value on substitution in the differentiable function.
$\therefore$ The values of ‘x’ for which the function ‘f’ is differentiable for $f(x) = \dfrac{2}{{(x - 3)}}$ are all numbers except for the value 3.

Note: Many students make the mistake of writing the values of x for which the function will be differentiable as 0 as they differentiate the function and equate the complete value of differentiation to 0. This solution gives the value 0. Keep in mind the word differentiable means the points where derivatives exist i.e. the points which on substitution in differentiation give some value.