Question

Find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$, if $y=(1+{{x}^{2}}){{\tan }^{-1}}x$.

Answer 1

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.