Answer
Verified
393.9k+ views
Hint: In order to solve this problem, we need to know the standard formulas of differentiation. The formulas are $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ and $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ . We need to find the first order and the second-order differentiation of the given function and substitute in the left-hand side expression to prove that right-hand side expression.
Complete step by step answer:
We are given the expression of $y={{\left( {{\tan }^{-1}}x \right)}^{2}}$ , and we need to prove that ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=2$ .
Let's differentiate the equation $y={{\left( {{\tan }^{-1}}x \right)}^{2}}$ with respect to x
We need to use the standard formulas such as $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ .
Performing the operation we get,
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( {{\tan }^{-1}}x \right)}^{2}} \\
& =2\left( {{\tan }^{-1}}x \right)\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right) \\
& =2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right)
\end{align}$
As we can see that in the expression that we have to prove we are given the second-order differential equation,
So, we need to differentiate the expression again.
The standard formula required are as follows,
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ .
By performing the differentiation with respect to x we get,
$\begin{align}
& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
& =2\left( \dfrac{-{{\tan }^{-1}}x\left( 2x \right)+\left( 1+{{x}^{2}} \right)\left( \dfrac{1}{1+{{x}^{2}}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \\
\end{align}$
Solving this we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)$
Now, substituting all the values from the expression we get,
$\begin{align}
& {{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx} \\
& {{\left( {{x}^{2}}+1 \right)}^{2}}\left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \right)+2x\left( {{x}^{2}}+1 \right)\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
\end{align}$
Simplifying the equation we get,
$\begin{align}
& {{\left( {{x}^{2}}+1 \right)}^{2}}\left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \right)+2x\left( {{x}^{2}}+1 \right)\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
& \left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{1} \right) \right)+2x\left( 2\left( {{\tan }^{-1}}x \right) \right) \\
& -4x{{\tan }^{-1}}x+2+4x{{\tan }^{-1}}x \\
& =2 \\
\end{align}$
Hence left-hand side = right-hand side.
Therefore, we can see that the ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=2$ is proved
Note:
The calculation in this problem is quite complicated. While differentiating we need to use the chain rule. The chain rule is used when we need to find the differentiation of the function inside the function. And it is carried out as we need to differentiate the outermost function followed by the multiplication of the differentiation of the innermost function.
Complete step by step answer:
We are given the expression of $y={{\left( {{\tan }^{-1}}x \right)}^{2}}$ , and we need to prove that ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=2$ .
Let's differentiate the equation $y={{\left( {{\tan }^{-1}}x \right)}^{2}}$ with respect to x
We need to use the standard formulas such as $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ .
Performing the operation we get,
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( {{\tan }^{-1}}x \right)}^{2}} \\
& =2\left( {{\tan }^{-1}}x \right)\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right) \\
& =2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right)
\end{align}$
As we can see that in the expression that we have to prove we are given the second-order differential equation,
So, we need to differentiate the expression again.
The standard formula required are as follows,
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ .
By performing the differentiation with respect to x we get,
$\begin{align}
& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
& =2\left( \dfrac{-{{\tan }^{-1}}x\left( 2x \right)+\left( 1+{{x}^{2}} \right)\left( \dfrac{1}{1+{{x}^{2}}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \\
\end{align}$
Solving this we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)$
Now, substituting all the values from the expression we get,
$\begin{align}
& {{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx} \\
& {{\left( {{x}^{2}}+1 \right)}^{2}}\left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \right)+2x\left( {{x}^{2}}+1 \right)\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
\end{align}$
Simplifying the equation we get,
$\begin{align}
& {{\left( {{x}^{2}}+1 \right)}^{2}}\left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right) \right)+2x\left( {{x}^{2}}+1 \right)\left( 2\left( {{\tan }^{-1}}x \right)\left( \dfrac{1}{1+{{x}^{2}}} \right) \right) \\
& \left( 2\left( \dfrac{-2x\left( {{\tan }^{-1}}x \right)+1}{1} \right) \right)+2x\left( 2\left( {{\tan }^{-1}}x \right) \right) \\
& -4x{{\tan }^{-1}}x+2+4x{{\tan }^{-1}}x \\
& =2 \\
\end{align}$
Hence left-hand side = right-hand side.
Therefore, we can see that the ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=2$ is proved
Note:
The calculation in this problem is quite complicated. While differentiating we need to use the chain rule. The chain rule is used when we need to find the differentiation of the function inside the function. And it is carried out as we need to differentiate the outermost function followed by the multiplication of the differentiation of the innermost function.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE