Question

If $x+y={{\tan }^{-1}}y\,and\,{{y}^{''}}=f\left( y \right){{y}^{'}}\,then\,f\left( y \right)=$
(a) $\dfrac{1}{y\left( 1+{{y}^{2}} \right)}$
(b) $\dfrac{3}{y\left( 1+{{y}^{2}} \right)}$
(c) $\dfrac{2}{y\left( 1+{{y}^{2}} \right)}$
(d) $\dfrac{-2}{y\left( 1+{{y}^{2}} \right)}$

Answer 1

We start solving the problem by differentiating both LHS and RHS with respect to x. We use the properties $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$ and $\dfrac{d}{dx}\left( f\left( y \right) \right)=\dfrac{d}{dy}f\left( y \right)\times \dfrac{dy}{dx}$ to proceed further through the problem. We then use $\dfrac{d\left( {{\tan }^{-1}}y \right)}{dy}=\dfrac{1}{1+{{y}^{2}}}$ and make necessary calculations and arrangements in the problem to get the desired result.

Complete step by step answer:
Given equation: $x+y={{\tan }^{-1}}y........(i)$
On differentiating both sides with respect to x-
\dfrac{d\left( x+y \right)}{dx}=\dfrac{d\left( {{\tan }^{-1}}y \right)}{dy}............(ii)$$$$$ Important property of differentiation- a. \dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$. b.$\dfrac{d}{dx}\left( f\left( y \right) \right)=\dfrac{d}{dy}f\left( y \right)\times \dfrac{dy}{dx}$. By using the property$\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)in equation (ii) - $$\dfrac{dx}{dx}+\dfrac{dy}{dx}=\dfrac{d\left( {{\tan }^{-1}}y \right)}{dx}.............(iii)$$ We will use standard notation of differentiation: \begin{aligned}
& y'=\dfrac{dy}{dx},and \\
& y''=\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \\
\end{aligned}$Now, further solving equation (iii), we get :$1+y'=\dfrac{d}{dx}\left( {{\tan }^{-1}}y \right)$By using property$\dfrac{d}{dx}\left( fy \right)=\dfrac{d}{dy}fy\times \dfrac{dy}{dx}$ , we get

& 1+{y’}=\dfrac{d}{dy}{{\tan }^{-1}}y\times \dfrac{dy}{dx} \\\ & \Rightarrow 1+{y’}=\dfrac{d}{dy}\left( {{\tan }^{-1}}y \right)\times y' \\\ \end{aligned}$$ We know that differentiation of $${{\tan }^{-1}}y$$ w.r.t $$y=\dfrac{1}{1+{{y}^{2}}}$$ i.e. $$\dfrac{d\left( {{\tan }^{-1}}y \right)}{dy}=\dfrac{1}{1+{{y}^{2}}}$$ Using this in above equation- $$1+y'=y'\times \left( \dfrac{1}{1+{{y}^{2}}} \right)............(iv)$$ Now, on again differentiating both sides with respect to x, we will get- $$\dfrac{d}{dx}\left( 1+y' \right)=\dfrac{d}{dx}\left( \dfrac{y'}{1+{{y}^{2}}} \right)............(v)$$ $$\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( y' \right)=\dfrac{d}{dx}\left\\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\\}$$ $$\begin{aligned} & \Rightarrow 0+\dfrac{d}{dx}\left( y' \right)=\dfrac{d}{dx}\left\\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\\} \\\ & \\\ \end{aligned}$$ Using standard notation conversation $$\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left\\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\\} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left\\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\\} \\\ & \Rightarrow y''=\dfrac{d}{dx}\left\\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\\} \\\ \end{aligned}$$ Now, we know that the product rule of differentiation is given as :$$\dfrac{d}{dx}\left\\{ f\left( x \right)\times g\left( x \right) \right\\}=f\left( x \right)\times \dfrac{d}{dx}g\left( x \right)+g\left( x \right)\times \dfrac{d}{dx}f\left( x \right)$$ $$\Rightarrow y''=y'\times \dfrac{d}{dx}\left( \dfrac{1}{1+{{y}^{2}}} \right)+\dfrac{1}{1+{{y}^{2}}}\times \dfrac{d}{dx}\left( y' \right)$$ $$\Rightarrow y''=y'\times \dfrac{d}{dy}\left( \dfrac{1}{1+{{y}^{2}}} \right)\times \dfrac{dy}{dx}+\dfrac{1}{1+{{y}^{2}}}\times \dfrac{d}{dx}\left( y' \right)$$ We know, $$\dfrac{d}{dy}\left( \dfrac{1}{1+{{y}^{2}}} \right)=\dfrac{d}{dy}{{\left( 1+{{y}^{2}} \right)}^{-1}}=-\left( 2y \right){{\left( 1+{{y}^{2}} \right)}^{-2}}=\dfrac{-1\left( 2y \right)}{{{\left( 1+{{y}^{2}} \right)}^{2}}}$$ . Putting this value in the above equation, we will get. $\Rightarrow y''=y'\times \left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right)+\dfrac{1}{1+{{y}^{2}}}\dfrac{d}{dx}\left( y' \right)$ We know $\begin{aligned} & \dfrac{d}{dx}\left( y' \right)=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=y'' \\\ & \\\ \end{aligned}$ $\begin{aligned} & \Rightarrow y''=y'\left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right)+\dfrac{1}{1+{{y}^{2}}}y'' \\\ & \\\ \end{aligned}$ Taking all the terms containing y’’ to one side of equation, we will get- $\begin{aligned} & \Rightarrow y''\left( 1-\dfrac{1}{1+{{y}^{2}}} \right)=y'\left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right) \\\ & \Rightarrow y''\left( \dfrac{1+{{y}^{2}}-1}{1+{{y}^{2}}} \right)=y'\left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right) \\\ \end{aligned}$ Dividing both sides of equation by y, we will get, $\Rightarrow y''\left[ \dfrac{y}{{{y}^{2}}+1} \right]=\dfrac{-2y'}{{{\left( 1+{{y}^{2}} \right)}^{2}}}$ Multiplying both sides of equation by $\left( 1+{{y}^{2}} \right)$ we will get- $\begin{aligned} & \Rightarrow y\times y''=\dfrac{-2y'}{{{\left( 1+{{y}^{2}} \right)}}} \\\ & \Rightarrow y''=\dfrac{-2}{y{{\left( 1+{{y}^{2}} \right)}}}y' \\\ \end{aligned}$ Given $y''=f\left( y \right)y'$, so $f\left( y \right)=\dfrac{-2}{y{{\left( 1+{{y}^{2}} \right)}}}\,$ **So, the correct answer is “Option D”.** **Note:** We should not make mistakes in signs, additions while performing differentiation operations. We can also solve the problem by differentiating both sides with respect to y for the first derivative to get $\dfrac{dx}{dy}$ and then take the inverse of it to get the equation for $y'$. We can then differentiate $y'$ in order to get $y''$. After finding both $y'$ and $y''$, we can compare both to get the required result.