Answer
Verified
425.4k+ views
Hint: Directly differentiate with respect to ‘x’ and remember the formula $\dfrac{d}{dx}({{\tan }^{-1}}x)=\dfrac{1}{1+{{x}^{2}}}$.
Complete step-by-step answer:
The given expression is,
\[y=(1+{{x}^{2}}){{\tan }^{-1}}x\]
Now we will find the first order derivative of the given expression, so we will differentiate the given expression with respect to $'x'$, we get
$\Rightarrow$ \[\dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left[ (1+{{x}^{2}}){{\tan }^{-1}}x \right]\]
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v\right) = u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{dy}{dx}=(1+{{x}^{2}})\dfrac{d}{dx}\left[ {{\tan }^{-1}}x \right]+{{\tan }^{-1}}x\dfrac{d}{dx}\left[ (1+{{x}^{2}}) \right]\]
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e., $\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$, applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{dy}{dx}=(1+{{x}^{2}})\dfrac{d}{dx}\left[ {{\tan }^{-1}}x \right]+{{\tan }^{-1}}x\left[\dfrac{d}{dx}(1)+\dfrac{d}{dx}\left[ {{x}^{2}} \right] \right]\]
Now we know the formula, $\dfrac{d}{dx}({{\tan }^{-1}}x)=\dfrac{1}{1+{{x}^{2}}}$, applying this formula, the above equation becomes,
$\Rightarrow$ \[\dfrac{dy}{dx}=(1+{{x}^{2}})\cdot \dfrac{1}{(1+{{x}^{2}})}+{{\tan }^{-1}}x\left[ \dfrac{d}{dx}(1)+\dfrac{d}{dx}\left[ {{x}^{2}} \right] \right]\]
Now we know $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ and derivative of a constant is always zero, applying this formula, the above equation becomes,
$\Rightarrow$ \[\dfrac{dy}{dx}=2x\cdot {{\tan }^{-1}}x+(1+{{x}^{2}})\cdot \dfrac{1}{(1+{{x}^{2}})}\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{dy}{dx}=2x\cdot {{\tan }^{-1}}x+1...........(i)\]
Now we will find the second order derivative. For that we will again differentiate the above
equation with respect to $'x'$, we get
$\Rightarrow$ \[\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( 2x\cdot {{\tan }^{-1}}x+1\right)\]
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e., $\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$, applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x\cdot {{\tan }^{-1}}x
\right)+\dfrac{d}{dx}\left( 1 \right)\]
Derivative of constant term is zero, so
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x\cdot {{\tan }^{-1}}x \right)+0\]
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v\right) = u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2x\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)+{{\tan }^{-1}}x\dfrac{d}{dx}(2x)\]
Now we know the formula, $\dfrac{d}{dx}({{\tan }^{-1}}x)=\dfrac{1}{1+{{x}^{2}}}$, applying this formula, the above equation becomes,
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2{{\tan }^{-1}}x+2x\cdot\dfrac{1}{1+{{x}^{2}}}\]
Taking the LCM, we get
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2(1+{{x}^{2}}){{\tan }^{-1}}x+2x}{1+{{x}^{2}}}\]
But from the given expression we have, \[y=(1+{{x}^{2}}){{\tan }^{-1}}x\], substituting this value, the above equation becomes,
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\left( y+x \right)}{1+{{x}^{2}}}\]
This is the required answer.
Note: In differentiation questions we should carefully observe the question ask for differentiation with respect to which variable, sometimes its with respect to $x$ and sometimes it is with respect to $'y'$ . As in both cases we will get different answers.
Complete step-by-step answer:
The given expression is,
\[y=(1+{{x}^{2}}){{\tan }^{-1}}x\]
Now we will find the first order derivative of the given expression, so we will differentiate the given expression with respect to $'x'$, we get
$\Rightarrow$ \[\dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left[ (1+{{x}^{2}}){{\tan }^{-1}}x \right]\]
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v\right) = u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{dy}{dx}=(1+{{x}^{2}})\dfrac{d}{dx}\left[ {{\tan }^{-1}}x \right]+{{\tan }^{-1}}x\dfrac{d}{dx}\left[ (1+{{x}^{2}}) \right]\]
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e., $\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$, applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{dy}{dx}=(1+{{x}^{2}})\dfrac{d}{dx}\left[ {{\tan }^{-1}}x \right]+{{\tan }^{-1}}x\left[\dfrac{d}{dx}(1)+\dfrac{d}{dx}\left[ {{x}^{2}} \right] \right]\]
Now we know the formula, $\dfrac{d}{dx}({{\tan }^{-1}}x)=\dfrac{1}{1+{{x}^{2}}}$, applying this formula, the above equation becomes,
$\Rightarrow$ \[\dfrac{dy}{dx}=(1+{{x}^{2}})\cdot \dfrac{1}{(1+{{x}^{2}})}+{{\tan }^{-1}}x\left[ \dfrac{d}{dx}(1)+\dfrac{d}{dx}\left[ {{x}^{2}} \right] \right]\]
Now we know $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ and derivative of a constant is always zero, applying this formula, the above equation becomes,
$\Rightarrow$ \[\dfrac{dy}{dx}=2x\cdot {{\tan }^{-1}}x+(1+{{x}^{2}})\cdot \dfrac{1}{(1+{{x}^{2}})}\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{dy}{dx}=2x\cdot {{\tan }^{-1}}x+1...........(i)\]
Now we will find the second order derivative. For that we will again differentiate the above
equation with respect to $'x'$, we get
$\Rightarrow$ \[\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( 2x\cdot {{\tan }^{-1}}x+1\right)\]
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e., $\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$, applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x\cdot {{\tan }^{-1}}x
\right)+\dfrac{d}{dx}\left( 1 \right)\]
Derivative of constant term is zero, so
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x\cdot {{\tan }^{-1}}x \right)+0\]
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v\right) = u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2x\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)+{{\tan }^{-1}}x\dfrac{d}{dx}(2x)\]
Now we know the formula, $\dfrac{d}{dx}({{\tan }^{-1}}x)=\dfrac{1}{1+{{x}^{2}}}$, applying this formula, the above equation becomes,
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2{{\tan }^{-1}}x+2x\cdot\dfrac{1}{1+{{x}^{2}}}\]
Taking the LCM, we get
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2(1+{{x}^{2}}){{\tan }^{-1}}x+2x}{1+{{x}^{2}}}\]
But from the given expression we have, \[y=(1+{{x}^{2}}){{\tan }^{-1}}x\], substituting this value, the above equation becomes,
$\Rightarrow$ \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\left( y+x \right)}{1+{{x}^{2}}}\]
This is the required answer.
Note: In differentiation questions we should carefully observe the question ask for differentiation with respect to which variable, sometimes its with respect to $x$ and sometimes it is with respect to $'y'$ . As in both cases we will get different answers.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
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
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE