Question

How do you differentiate ${{e}^{\dfrac{1}{2x}}}$?

Answer 1

Hint: Now the given function is a composite function. Hence to differentiate the function we will have to use chain rule of differentiation. Now we know that according to chain rule of differentiation we have the differentiation of $f\left( g\left( x \right) \right)$ is given by $f'\left( g\left( x \right) \right).g'\left( x \right)$ . Now we know that the differentiation of ${{e}^{x}}$ is ${{e}^{x}}$ and the differentiation of $\dfrac{1}{2x}$ is $\dfrac{-1}{2{{x}^{2}}}$ . Hence we will use this rule and find the differentiation of ${{e}^{\dfrac{1}{2x}}}$

Complete step by step solution:
Now consider the given function ${{e}^{\dfrac{1}{2x}}}$ .
Now we can see that the given function is a composite function. To differentiate composite functions we use chain rule of differentiation.
Let us understand how to differentiate using the chain rule.
Now let us say that we have a composite function $f\left( g\left( x \right) \right)$
Now differentiation of such function by chain rule is given by $f'\left( g\left( x \right) \right).g'\left( x \right)$ .
Now let us use this rule to differentiate the given function.
Let us consider ${{e}^{\dfrac{1}{2x}}}$
Now here we have two functions $f\left( x \right)={{e}^{x}}$ and $g\left( x \right)=\dfrac{1}{2x}$ .
Now differentiating both functions we get, $f'\left( x \right)={{e}^{x}}$ and $g'\left( x \right)=-\dfrac{1}{2{{x}^{2}}}$ .
Now substituting the values in the formula $f'\left( g\left( x \right) \right).g'\left( x \right)$ we get
$\Rightarrow \dfrac{d\left( {{e}^{\dfrac{1}{2x}}} \right)}{dx}=-{{e}^{\dfrac{1}{2x}}}.\dfrac{1}{2{{x}^{2}}}$
Hence the we get the differentiation of the given function is $-\dfrac{{{e}^{\dfrac{1}{2x}}}}{2{{x}^{2}}}$ .

Note: Now note that we can use chain rule for series of composite functions. Now note that in chain rule we do not have a product of two functions but the composition of two functions. Composition means the input of outer functions we have the output of inner functions. For multiplication of functions we have $\left( uv \right)'=u'v+v'u$ . Hence not to be confused in uv rule and the chain rule of differentiation.