Question

If $y=\sqrt{x}+\dfrac{1}{\sqrt{x}}$, prove that $2x\dfrac{dy}{dx}+y=2\sqrt{x}$.

Answer 1

Hint: For solving this question we will first square the terms on the left-hand side and right-hand sides in the given equation. Then, we will try to get the equation $x{{y}^{2}}={{x}^{2}}+1+2x$. After that, we will apply the product rule of differential calculus and formulas like $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, $\dfrac{d\left( {{y}^{n}} \right)}{dx}=\left( n{{y}^{n-1}} \right)\dfrac{dy}{dx}$. Then, we will arrange the terms in the result to prove the desired result easily.

Complete step-by-step answer:
It is given that, $y=\sqrt{x}+\dfrac{1}{\sqrt{x}}$ and we have to prove the following equation:
$2x\dfrac{dy}{dx}+y=2\sqrt{x}$
Now, before we proceed, we should know the following formula:
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.............\left( 1 \right)$
Now, we will square the terms on the left-hand side and right-hand sides in the given equation. Then,
$\begin{align}
  & y=\sqrt{x}+\dfrac{1}{\sqrt{x}} \\
 & \Rightarrow {{y}^{2}}={{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}} \\
\end{align}$
Now, we will use the formula from the equation to write ${{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}}=x+\dfrac{1}{x}+2$ in the above equation. Then,
$\begin{align}
  & {{y}^{2}}={{\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)}^{2}} \\
 & \Rightarrow {{y}^{2}}={{\left( \sqrt{x} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{x}} \right)}^{2}}+2\times \sqrt{x}\times \dfrac{1}{\sqrt{x}} \\
 & \Rightarrow {{y}^{2}}=x+\dfrac{1}{x}+2 \\
\end{align}$
Now, we will multiply the above equation by the variable $x$ . Then,
$\begin{align}
  & {{y}^{2}}=x+\dfrac{1}{x}+2 \\
 & \Rightarrow x{{y}^{2}}={{x}^{2}}+x\times \dfrac{1}{x}+2x \\
 & \Rightarrow x{{y}^{2}}={{x}^{2}}+1+2x...........................\left( 2 \right) \\
\end{align}$
Now, we will use the formula from the equation (1) to write ${{x}^{2}}+1+2x={{\left( x+1 \right)}^{2}}$ in the above equation. Then,
$\begin{align}
  & x{{y}^{2}}={{x}^{2}}+1+2x \\
 & \Rightarrow x{{y}^{2}}={{\left( x+1 \right)}^{2}} \\
\end{align}$
Now, we will take the square-root of the terms on the left-hand side and right-hand sides in the above equation. Then,
$\begin{align}
  & x{{y}^{2}}={{\left( x+1 \right)}^{2}} \\
 & \Rightarrow \sqrt{x{{y}^{2}}}=\sqrt{{{\left( x+1 \right)}^{2}}} \\
 & \Rightarrow y\sqrt{x}=\left( x+1 \right)........................\left( 3 \right) \\
\end{align}$
Now, before we proceed, we should know the following formulas and concepts of differential calculus:

1. If $y=f\left( x \right)\cdot g\left( x \right)$ , then $\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right)\cdot g\left( x \right) \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right)$. This is also known as the product rule of differentiation.

2. If $y={{x}^{n}}$ , then $\dfrac{dy}{dx}=\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$.

3. Derivative of ${{y}^{n}}$ with respect to $x$ will be $\dfrac{d\left( {{y}^{n}} \right)}{dx}=\left( n{{y}^{n-1}} \right)\dfrac{dy}{dx}$ .
Now, from the equation (2) we have the following equation:
$x{{y}^{2}}={{x}^{2}}+1+2x$
Now, we will differentiate the above equation with respect to $x$ . Then,
$\begin{align}
  & x{{y}^{2}}={{x}^{2}}+1+2x \\
 & \Rightarrow \dfrac{d\left( x{{y}^{2}} \right)}{dx}=\dfrac{d\left( {{x}^{2}}+1+2x \right)}{dx} \\
\end{align}$
Now, we will use the product rule of differentiation to write $\dfrac{d\left( x{{y}^{2}} \right)}{dx}={{y}^{2}}+2xy\dfrac{dy}{dx}$ in the above equation. Then,
$\begin{align}
  & \dfrac{d\left( x{{y}^{2}} \right)}{dx}=\dfrac{d\left( {{x}^{2}}+1+2x \right)}{dx} \\
 & \Rightarrow \dfrac{d\left( x \right)}{dx}\times {{y}^{2}}+x\times \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( {{x}^{2}} \right)}{dx}+\dfrac{d\left( 1 \right)}{dx}+\dfrac{d\left( 2x \right)}{dx} \\
\end{align}$
Now, we will use the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ to write $\dfrac{d\left( x \right)}{dx}=1$ , $\dfrac{d\left( {{x}^{2}} \right)}{dx}=2x$ , $\dfrac{d\left( 2x \right)}{dx}=2$ and formula $\dfrac{d\left( {{y}^{n}} \right)}{dx}=\left( n{{y}^{n-1}} \right)\dfrac{dy}{dx}$ to write $\dfrac{d\left( {{y}^{2}} \right)}{dx}=2y\dfrac{dy}{dx}$ and as $1$ is a constant term so, we can write $\dfrac{d\left( 1 \right)}{dx}=0$ in the above equation. Then,
$\begin{align}
  & \dfrac{d\left( x \right)}{dx}\times {{y}^{2}}+x\times \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( {{x}^{2}} \right)}{dx}+\dfrac{d\left( 1 \right)}{dx}+\dfrac{d\left( 2x \right)}{dx} \\
 & \Rightarrow {{y}^{2}}+x\times 2y\dfrac{dy}{dx}=2x+0+2 \\
 & \Rightarrow {{y}^{2}}+2xy\dfrac{dy}{dx}=2\left( x+1 \right) \\
 & \Rightarrow y\left( y+2x\dfrac{dy}{dx} \right)=2\left( x+1 \right) \\
 & \Rightarrow y+2x\dfrac{dy}{dx}=\dfrac{2\left( x+1 \right)}{y} \\
\end{align}$
Now, we will substitute $\left( x+1 \right)=y\sqrt{x}$ from equation (3) in the above equation. Then,
$\begin{align}
  & y+2x\dfrac{dy}{dx}=\dfrac{2\left( x+1 \right)}{y} \\
 & \Rightarrow y+2x\dfrac{dy}{dx}=\dfrac{2y\sqrt{x}}{y} \\
 & \Rightarrow y+2x\dfrac{dy}{dx}=2\sqrt{x} \\
 & \Rightarrow 2x\dfrac{dy}{dx}+y=2\sqrt{x} \\
\end{align}$
Now, from the above result, we conclude that, $2x\dfrac{dy}{dx}+y=2\sqrt{x}$ .
Hence proved.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to prove the desired result quickly. And to avoid long calculations, we should first try to get an equation in $x$ and $y$ such that we will get a term like $x\dfrac{dy}{dx}$ after differentiating it. Moreover, we should apply the product rule of differential calculus, and formulas like $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, $\dfrac{d\left( {{y}^{n}} \right)}{dx}=\left( n{{y}^{n-1}} \right)\dfrac{dy}{dx}$ correctly without any mathematical error to prove the desired result easily.