Question

How do I find the derivative of the function $y=\log \left( {{x}^{2}}+1 \right)$?

Answer 1

Hint: We start solving the problem by assuming \[{{x}^{2}}+1=z\] and then differentiating both sides of the given function with respect to x. We then recall the chain rule of differentiation as $\dfrac{d\left( g\left( f \right) \right)}{dx}=\dfrac{d\left( g \right)}{df}\times \dfrac{df}{dx}$ to proceed through the problem. We then make use of the fact that $\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}$ to proceed through the problem. We then make use of the facts that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, $\dfrac{d\left( a \right)}{dx}=0$ to get the required answer for the derivative of the function.

Complete step by step answer:
According to the problem, we are asked to find the derivative of the function $y=\log \left( {{x}^{2}}+1 \right)$.
We have $y=\log \left( {{x}^{2}}+1 \right)$ ---(1).
Let us assume \[{{x}^{2}}+1=z\]. Let us substitute this in equation (1).
$\Rightarrow y=\log z$ ---(2).
Let us differentiate both sides of the equation (2) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \log z \right)}{dx}$ ---(3).
From chain rule of differentiation, we know that $\dfrac{d\left( g\left( f \right) \right)}{dx}=\dfrac{d\left( g \right)}{df}\times \dfrac{df}{dx}$. Let us substitute this result in equation (3).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \log z \right)}{dz}\times \dfrac{dz}{dx}$ ---(4).
We know that $\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}$. Let us use this result in equation (4).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{z}\times \dfrac{dz}{dx}$ ---(5).
Now, let us substitute $z={{x}^{2}}+1$ in equation (5).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\left( {{x}^{2}}+1 \right)}\times \dfrac{d\left( {{x}^{2}}+1 \right)}{dx}$ ---(6).
We know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, $\dfrac{d\left( a \right)}{dx}=0$. Let us use this result in equation (6).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\left( {{x}^{2}}+1 \right)}\times 2x$.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x}{{{x}^{2}}+1}$.
$\therefore $ We have found the derivative of the function $y=\log \left( {{x}^{2}}+1 \right)$ as $\dfrac{2x}{{{x}^{2}}+1}$.

Note:
Whenever we get this type of problem, we try to make use of chain rules to get a solution to the given problem. We should not forget to different ${{x}^{2}}+1$ after performing equation (5) which is the common mistake done by students. We can also solve this problem by making use of the fact that \[\dfrac{d}{dx}\left( \ln \left( f\left( x \right) \right) \right)=\dfrac{\dfrac{d\left( f\left( x \right) \right)}{dx}}{f\left( x \right)}\] to get the required answer. Similarly, we can expect problems to find the derivative of the function $y=\cos \left( \log \left( 8x \right) \right)$.