Question

How do I find the derivative of $\ln (\sqrt x )$ ?

Answer 1

Hint: In the question above, we have an equation $\ln (\sqrt x )$ , and we are supposed to find its derivative. One of the methods to find a derivative is using the chain rule, and since we have a square root involved, we will use this chain rule for derivatives.
The chain rule for derivatives is a rule that we use to find the derivative of functions of the form $f(g(x))$ .

Complete step-by-step solution:
The chain rule for derivatives has a formula:
$h(x) = f(g(x))$
And,
$h'(x) = f'(g(x)).g'(x)$
Now, to use the chain rule for derivatives, we need to first make sure that our equation satisfies the needs of that rule, so comparing the equation, we get,
$f(x) = \ln (x)$
And,
$g(x) = \sqrt x $
Then,
$f(g(x)) = \ln (\sqrt x )$
This tells us that we can use the chain rule for derivatives to find the derivative of $\ln (\sqrt x )$ .
To use the chain rule, we also have to find $h'(x) = f'(g(x)).g'(x)$ . Since in our example we have$f(x) = \ln (x)$ and $g(x) = \sqrt x $ , we need to find the other values, too.
The derivative of $\ln (x)$ is $\dfrac{1}{x}$
The derivative of $\sqrt x $ is $(\dfrac{1}{2}){x^{(\dfrac{{ - 1}}{2})}}$ or $\dfrac{1}{{2\sqrt x }}$
Since we have the derivative of $\ln (x)$ as $\dfrac{1}{x}$ , we have that $f'(x) = \dfrac{1}{x}$ , so $f'(g(x)) = \dfrac{1}{{\sqrt x }}$ .
Also, we know that, $g'(x) = \dfrac{1}{{2\sqrt x }}$ .
Putting all the values in the chain rule for derivatives,
$ \Rightarrow f(x) = \ln (x)$
$ \Rightarrow g(x) = \sqrt x $
And,
$ \Rightarrow h(x) = f(g(x)) = \ln (\sqrt x )$
Now we know that,
$ \Rightarrow h'(x) = f'(g(x)).g'(x)$
Putting in the values of $f'(g(x)) = \dfrac{1}{{\sqrt x }}$ and $g'(x) = \dfrac{1}{{2\sqrt x }}$
Therefore,
$ \Rightarrow h'(x) = \dfrac{1}{{\sqrt x }}.\dfrac{1}{{2\sqrt x }}$
Simplifying the equation,
$ \Rightarrow h'(x) = \dfrac{1}{{2x}}$

The derivative of $\ln (\sqrt x )$ is $\dfrac{1}{{2x}}$ .

Note: The chain rule for derivatives is mainly used for composite functions. A composite function is a function that can be constructed as $f(g(x))$ . We use it when we have to differentiate the function of a function. We use the product rule when differentiating two functions, multiplied together.