Question

Differentiate the given function: \[f\left( x \right)=\log \left( x+\sqrt{1+{{x}^{2}}} \right)\]

Answer 1

Hint: First we will write the function using $\sqrt[n]{{{a}^{x}}}={{a}^{\dfrac{x}{n}}}$ , then we will apply the chain rule to start with our differentiation that is $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right).g'\left( x \right)$ , once we get this we will apply the differentiation of logarithm that is $f\left( t \right)=\log t\Rightarrow f'\left( t \right)=\dfrac{1}{t}$ to differentiate the first term. For the second term first we will expand it using the sum rule that is $\dfrac{d\left[ f\left( x \right)+g\left( x \right) \right]}{dx}=f'\left( x \right)+g'\left( x \right)$ then we will use the chain rule as we saw above and eventually power rule $\dfrac{d}{dx}\left[ {{x}^{n}} \right]=n{{x}^{n-1}}$. Finally, we will simplify the expression and get the answer.

Complete step by step answer:
It is given that we have to differentiate the function: \[f\left( x \right)=\log \left( x+\sqrt{1+{{x}^{2}}} \right)\]
Now first we will use $\sqrt[n]{{{a}^{x}}}={{a}^{\dfrac{x}{n}}}$ to rewrite $\sqrt{1+{{x}^{2}}}={{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}$
So our function will become: \[f\left( x \right)=\log \left( x+\left( 1+{{x}^{2}} \right)\dfrac{1}{2} \right)\]
We will start by using chain rule which states that: $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right).g'\left( x \right)$
Now, we have with us the function: $\dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d}{dx}\left[ \log \left( x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right) \right]$ , we will apply chain rule on this so according to chain here, we have $f\left( x \right)=\log \left( g\left( x \right) \right)\text{ and }g\left( x \right)=x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}$ , now we know that if $f\left( t \right)=\log t\Rightarrow f'\left( t \right)=\dfrac{1}{t}$ , therefore $f'\left( g\left( x \right) \right)=\dfrac{1}{g\left( x \right)}$ ,


Putting these values in our function:
\[\begin{align}
  & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d}{dx}\left[ \log \left( x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right) \right] \\
 & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{1}{\left( x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)}\times \dfrac{d}{dx}\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]\text{ }.......\text{Equation 1} \\
\end{align}\]
Now, we will differentiate the second term of the given function that is $x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}$ , we know that the sum rule in derivative states that : $\dfrac{d\left[ f\left( x \right)+g\left( x \right) \right]}{dx}=f'\left( x \right)+g'\left( x \right)$ , therefore
$\dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=\dfrac{d}{dx}\left[ x \right]+\dfrac{d}{dx}\left[ {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]$
Now, we know the power rule states that $\dfrac{d}{dx}\left[ {{x}^{n}} \right]=n{{x}^{n-1}}$ , therefore:
$\begin{align}
  & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=\dfrac{d}{dx}\left[ x \right]+\dfrac{d}{dx}\left[ {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right] \\
 & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1{{x}^{1-1}}+\dfrac{d}{dx}\left[ {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right] \\
 & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}={{x}^{0}}+\dfrac{d}{dx}\left[ {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right] \\
 & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1+\dfrac{d}{dx}\left[ {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right] \\
\end{align}$

We will now again use the chain rule which states that: $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right).g'\left( x \right)$ :
Here, $f\left( x \right)={{\left( g\left( x \right) \right)}^{\dfrac{1}{2}}}\text{ and g}\left( x \right)=\left( 1+{{x}^{2}} \right)$ , we will apply power rule as we saw above on each of these functions therefore:
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1+\dfrac{d}{dx}\left[ {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right] \\
 & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1+\dfrac{1}{2}\left[ {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}-1}} \right]\dfrac{d}{dx}\left[ \left( 1+{{x}^{2}} \right) \right] \\
\end{align}\]
Now, we will expand the $\dfrac{d}{dx}$ into the bracket, therefore we will get:
\[\Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1+\dfrac{1}{2}\left[ {{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}} \right]\left( \dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right) \right)\]
We will now differentiate the terms and after differentiating we will get:
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1+\dfrac{1}{2}\left[ {{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}} \right]\left( 0+2\left( {{x}^{2-1}} \right) \right) \\
 & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1+\dfrac{1}{2{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}\left( 2x \right) \\
 & \Rightarrow \dfrac{d\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right]}{dx}=1+\dfrac{x}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}} \\
\end{align}\]

We will now put this value in equation 1 that is:
\[\begin{align}
  & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{1}{\left( x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)}\times \dfrac{d}{dx}\left[ x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right] \\
 & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{1}{\left( x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)}\times \left( 1+\dfrac{x}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}} \right) \\
\end{align}\]
Now, we will see simplify the expression, first by adding the terms in bracket:
\[\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{1}{\left( x+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)}\times \left( \dfrac{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}+x}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}} \right)\]
We will now cancel the common terms:
\[\begin{align}
  & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{1}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}} \\
 & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{1}{\sqrt{1+{{x}^{2}}}} \\
\end{align}\]
Hence, the differentiation of \[f\left( x \right)=\log \left( x+\sqrt{1+{{x}^{2}}} \right)\] is $f'\left( x \right)=\dfrac{1}{\sqrt{1+{{x}^{2}}}}$ .

Note:
 While applying $\sqrt[n]{{{a}^{x}}}={{a}^{\dfrac{x}{n}}}$ to rewrite $\sqrt{1+{{x}^{2}}}={{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}$, we by default took $n=2$ as it is understood that we are taking the square root whenever no number is mentioned but let’s suppose if it was written like: $\sqrt[3]{\left( 1+{{x}^{2}} \right)}$ then it would have meant a cube root and we would have taken$n=3$. As there are lots of calculation involved be careful while writing the equations down, one small mistake can change the whole answer.