Question

How do you find the derivative of $y = \ln (\sqrt x )$?

Answer 1

Hint:We first define the chain rule and how the differentiation of composite function works. We take differentiation of the main function with respect to an intermediate function and then take differentiation of the intermediate function with respect to $x$. At last, we multiply both the terms to get the final result.

Complete step by step solution:
(i)
As we are given $y = \ln (\sqrt x )$, we will differentiate the given function with respect to $x$ using the chain rule.

Here, we have a composite function where the main function is $g(x) = \ln x$ and the other function is

$h(x) = \sqrt x $
We have $goh(x) = g(\sqrt x ) = \ln (\sqrt x )$. We take this as our given function $f(x) = \ln (\sqrt x )$

(ii)
Now, we need to find the value of $\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] =

\dfrac{d}{{dx}}\left[
{\ln \left( {\sqrt x } \right)} \right]$.
As we know that,
$f\left( x \right) = goh\left( x \right)$

Differentiating both the sides, we will get:
$\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {goh\left( x \right)} \right]$

Applying the chain rule here, we will get:
\[\dfrac{d}{{dx}}\left[ {goh\left( x \right)} \right] = \dfrac{d}{{d\left[ {h\left( x \right)} \right]}}\left[
{goh\left( x \right)} \right] \times \dfrac{{d\left[ {h\left( x \right)} \right]}}{{dx}}\]

Writing it in other words,
$\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = g'\left[ {h\left( x \right)} \right]h'\left( x \right)$

The chain rule allows us to differentiate with respect to the function $h\left( x \right)$ instead of $x$
and after that, we need to take the differentiated form of $h\left( x \right)$ with respect to $x$.

For function $f\left( x \right) = \ln \left( {\sqrt x } \right)$, we take differentiation of $f\left( x \right) =
\ln \left( {\sqrt x } \right)$ with respect to the function $h(x) = \sqrt x $ instead of $x$ and after that we need to take the differentiated form of $h(x) = \sqrt x $ with respect to $x$.

(iii)
As we know that,
$\dfrac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{\left( {n - 1} \right)}}$

Therefore, for $h(x) = \sqrt x $ , which can also be written as $h\left( x \right) = {x^{\dfrac{1}{2}}}$
differentiating both the sides, we will get:
$
h'\left( x \right) = \dfrac{1}{2}{x^{\left( {\dfrac{1}{2} - 1} \right)}} \\
h'\left( x \right) = \dfrac{1}{2}{x^{\dfrac{{ - 1}}{2}}} \\
h'\left( x \right) = \dfrac{1}{{2\sqrt x }} \\
$

And, we also know that since,
$g(x) = \ln x$

Differentiating both the sides,
$g'\left( x \right) = \dfrac{1}{x}$

(iv)
According to the chain rule, we have:
$\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{d\left[ {\sqrt x } \right]}}\left[ {\ln \left(
{\sqrt x } \right)} \right] \times \dfrac{{d\left[ {\sqrt x } \right]}}{{dx}}$

Putting the values, we calculated in the chain rule,
$\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{1}{{\sqrt x }} \times \dfrac{1}{{2\sqrt x }}$

On simplifying it further, we get:
$\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{1}{{2x}}$

Hence, the differentiation of $y = \ln (\sqrt x )$ is $\dfrac{1}{{2x}}$.

Note: We need to remember that in the chain rule $\dfrac{d}{{d\left[ {h\left( x \right)} \right]}}\left[ {goh\left( x \right)} \right] \times \dfrac{{d\left[ {h\left( x \right)} \right]}}{{dx}}$, we are not cancelling out the part $d\left[ {h\left( x \right)} \right]$. Cancellation of the base of differentiation is never possible.
It is just a notation to understand the function which is used as a base to differentiate.