Question

How do you differentiate $\ln {{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}$?

Answer 1

Hint: Now we are given with a composite function. Hence we will use chain rule of differentiation to solve the differential. Now using chain rule of differentiation we have differentiation of the function $f\left( g\left( h\left( x \right) \right) \right)$ as $f'\left( g\left( h\left( x \right) \right) \right)g'\left( h\left( x \right) \right).h'\left( x \right)$. Now we will find the differentiation of each function and hence substitute the values to find the differentiation of the given equation. To do so we have $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$, $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ and $\dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{1}{2\sqrt{x}}$

Complete step-by-step solution:
Now consider the given function $\ln {{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}$.
The given function is a composite function of the form $f\left( g\left( h\left( x \right) \right) \right)$ .
Now we know that to differentiate a composite function we use chain rule of differentiation.
Now using chain rule of differentiation we have $\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right)$ where $f'\left( x \right)=\dfrac{d\left( f\left( x \right) \right)}{dx}$ .
Now in the above function we have $f\left( x \right)=\ln x$, $g\left( x \right)=={{x}^{\dfrac{1}{2}}}$ and $h\left( x \right)={{x}^{2}}+1$.
Now consider $f\left( x \right)=\ln x$
Now we know that the differentiation of $\ln x$ is given by $\dfrac{1}{x}$ .
Hence we have $f'\left( x \right)=\dfrac{1}{x}$ .
Now consider $g\left( x \right)={{x}^{\dfrac{1}{2}}}$
Now differentiating we get, $g'\left( x \right)=\dfrac{1}{2{{x}^{\dfrac{1}{2}}}}$
And since $h\left( x \right)={{x}^{2}}+1$ we have $h'\left( x \right)=2x$ .
Now according to chain rule we have differentiation of the function $f\left( g\left( h\left( x \right) \right) \right)$ as
$\Rightarrow f'\left( g\left( h\left( x \right) \right) \right)g'\left( h\left( x \right) \right).h'\left( x \right)$
Hence on substituting the values in the given formula we have,
$\Rightarrow \dfrac{1}{{{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}}.\dfrac{1}{2{{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}}.\left( 2x \right)$
Now we know that ${{x}^{m}}{{x}^{n}}={{x}^{m+n}}$ . Hence using this and simplifying we get,
$\Rightarrow \dfrac{x}{{{x}^{2}}+1}$
Hence the differentiation of the given function is given by $\dfrac{x}{{{x}^{2}}+1}$.

Note: Now note that we can use chain rule for a complete series of composition of function. The formula for chain rule is $\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right)$. Hence when we have a series of compositions we will use this formula inductively to find the formula for n number of compositions. Hence we get differentiation of the function $f\left( g\left( h\left( x \right) \right) \right)$ as
$f'\left( g\left( h\left( x \right) \right) \right)g'\left( h\left( x \right) \right).h'\left( x \right)$ . Also note that composition is not multiplication of the functions.