Question

If we are given that \[y=\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]\], then show that \[\left( {{x}^{2}}+{{a}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}=0\]

Answer 1

Hint: To solve this question we will first find the values of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\,\,and\,\dfrac{dy}{dx}\] then put it in the expression that we have to prove. After that we will solve the obtained expression further and will prove it equal to the RHS of the equation i.e. 0. We need to know here that the expressions\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\,\,and\,\dfrac{dy}{dx}\] represents the double differentiation and differentiation of y with respect to x respectively. We will also be applying the chain rule and product rule to solve this question, the chain rule says, the derivative of f(g(x)) = f'(g(x)).g'(x). also we will use these formulas $\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x},\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$.
Complete step by step answer:
we are given that,
\[y=\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]\]
And we have to prove that,
\[\left( {{x}^{2}}+{{a}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}=0\,\,....\left( 1 \right)\]
First, we need to find the values of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\,\,and\,\dfrac{dy}{dx}\] in order to solve the above equation, so
\[y=\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]\]
Differentiating both sides with respect to x and applying chain rule on the RHS of the equation we get, we get
According to chain rule, the derivative of f(g(x)) = f'(g(x)).g'(x).

$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d\left( \log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right] \right)}{dx} \\
 & \dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}.\left( 1+\dfrac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right) \\
\end{align}$

Taking LCM, we get
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}.\left( \dfrac{\sqrt{{{x}^{2}}+{{a}^{2}}}+x}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right) \\
 & \\
\end{align}\]
\[\begin{align}
  & \dfrac{dy}{dx}=\left( \dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right) \\
 & \\
\end{align}\]
Multiplying by $x$on both sides, we get
\[\begin{align}
  & x.\dfrac{dy}{dx}=\left( \dfrac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right)\,\,\,....\left( 2 \right) \\
 & \\
\end{align}\]
Now again differentiating the \[\begin{align}
  & \dfrac{dy}{dx}=\left( \dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right) \\
 & \\
\end{align}\] with respect to $x$, we get
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right) \\
 & \\
\end{align}\]
Applying chain rule on the RHS of the equation, we get
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{1}{2}\left( \dfrac{2x}{{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{-\dfrac{3}{2}}}} \right) \\
 & \\
\end{align}\]
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{-x}{{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{-\dfrac{3}{2}}}} \right) \\
 & \\
\end{align}\]
Multiplying both sides by \[\left( {{x}^{2}}+{{a}^{2}} \right)\], we get
\[\left( {{x}^{2}}+{{a}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{-x}{\sqrt{\left( {{x}^{2}}+{{a}^{2}} \right)}} \right)\,\,\,....\left( 3 \right)\]
Putting the values of equation 2 and 3 in the equation 1, we get
\[\left( {{x}^{2}}+{{a}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}\]
\[\begin{align}
  & =\left( \dfrac{-x}{\sqrt{\left( {{x}^{2}}+{{a}^{2}} \right)}} \right)\,+\left( \dfrac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right)\, \\
 & =0 \\
\end{align}\]
= R. H. S.

Note: Whenever we are given a particular expression to prove then try to manipulate the given conditions such that we can prove the expression easily. And also read and learn basic properties of logarithmic functions in order to solve problems involving them. Also learn about the properties about differentiation related to logarithmic functions. And remember differentiation of $\log x$ is given by $\dfrac{1}{x}$.