Question

What is the derivative of $\sqrt{{{x}^{2}}+1}$?

Answer 1

Assume the function ${{x}^{2}}+1$ as $f\left( x \right)$ and write $\sqrt{{{x}^{2}}+1}$ as $\sqrt{f\left( x \right)}$. Now, use the chain rule of differentiation to differentiate the function. First differentiate the function $\sqrt{f\left( x \right)}$ with respect to the function $f\left( x \right)$ and then differentiate the function $f\left( x \right)$ with respect to x. Finally we will take the product of these two derivatives to get the answer. The formula is given as $\dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{dx}=\dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{d\left[ f\left( x \right) \right]}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$ . Use the formula $\dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{d\left[ f\left( x \right) \right]}=\dfrac{1}{2\sqrt{f\left( x \right)}}$ to simplify the first part.

Complete step-by-step solution:
Here we have been provided with the function $\sqrt{{{x}^{2}}+1}$ and we are asked to find its derivative. Here we will use the chain rule of derivative to get the answer. Assuming the function ${{x}^{2}}+1$ as $f\left( x \right)$ we have the function $\sqrt{{{x}^{2}}+1}$ of the form $\sqrt{f\left( x \right)}$. So we have,
$\Rightarrow \sqrt{{{x}^{2}}+1}=\sqrt{f\left( x \right)}$
On differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{d\left( \sqrt{{{x}^{2}}+1} \right)}{dx}=\dfrac{d\left( \sqrt{f\left( x \right)} \right)}{dx}$
Now, according to the chain rule of derivative first we have to differentiate the function $\sqrt{f\left( x \right)}$ with respect to $f\left( x \right)$ and then we have to differentiate $f\left( x \right)$ with respect to x. Finally, we need to consider their product to get the relation. So we get,
$\Rightarrow \dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{dx}=\dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{d\left[ f\left( x \right) \right]}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$
The derivative of $\sqrt{f\left( x \right)}$ with respect to $f\left( x \right)$ is given by the formula $\dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{d\left[ f\left( x \right) \right]}=\dfrac{1}{2\sqrt{f\left( x \right)}}$, so we get,
$\Rightarrow \dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{dx}=\dfrac{1}{2\sqrt{f\left( x \right)}}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$
Substituting the value of $f\left( x \right)$ we get,
$\begin{aligned} & \Rightarrow \dfrac{d\left[ \sqrt{{{x}^{2}}+1} \right]}{dx}=\dfrac{1}{2\sqrt{{{x}^{2}}+1}}\times \dfrac{d\left[ {{x}^{2}}+1 \right]}{dx} \\\ & \Rightarrow \dfrac{d\left[ \sqrt{{{x}^{2}}+1} \right]}{dx}=\dfrac{1}{2\sqrt{{{x}^{2}}+1}}\times \left( \dfrac{d\left[ {{x}^{2}} \right]}{dx}+\dfrac{d\left[ 1 \right]}{dx} \right) \\\ \end{aligned}$
Using the formula $\dfrac{d\left[ {{x}^{n}} \right]}{dx}=n{{x}^{n-1}}$ and the fact that the derivative of a constant is 0 we get,
$\begin{aligned} & \Rightarrow \dfrac{d\left[ \sqrt{{{x}^{2}}+1} \right]}{dx}=\dfrac{1}{2\sqrt{{{x}^{2}}+1}}\times \left( 2{{x}^{2-1}}+0 \right) \\\ & \Rightarrow \dfrac{d\left[ \sqrt{{{x}^{2}}+1} \right]}{dx}=\dfrac{1}{2\sqrt{{{x}^{2}}+1}}\times \left( 2x \right) \\\ & \therefore \dfrac{d\left[ \sqrt{{{x}^{2}}+1} \right]}{dx}=\dfrac{x}{\sqrt{{{x}^{2}}+1}} \\\ \end{aligned}$
Hence, the above relation is our answer.

Note: You must remember all the basic rules and formulas of differentiation like: - product rule, chain rule, $\dfrac{u}{v}$ rule etc. You can also convert the given function into parametric form then differentiate. Assume the function as y and substitute $x=\tan \theta$ and then you will get the function $y=\sec \theta$. Now, find the derivative by using the formula $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta } \right)}{\left( \dfrac{dx}{d\theta } \right)}$. Use the formulas $\dfrac{d\left( \sec \theta \right)}{dx}=\sec \theta \tan \theta$ and $\dfrac{d\left( \tan \theta \right)}{dx}={{\sec }^{2}}\theta$ to simplify and get the answer. Finally substitute the value of these trigonometric functions in terms of x to get the answer.