Question

If the given expression $y=\dfrac{1-{{x}^{4}}}{1+{{x}^{4}}}$, then $\dfrac{dy}{dx}.\dfrac{dx}{dy}$ is equal to
(a) $1$
(b) $xy$
(c) Does not exist
(d) $\dfrac{x+y}{xy}$

Answer 1

Hint: First find derivative with respect to $'x'$ and then derivative with respect to $'y'$ . Multiply both to get the result.

Complete step-by-step answer:
The given expression is $y=\dfrac{1-{{x}^{4}}}{1+{{x}^{4}}}$.
First, we shall find $\dfrac{dy}{dx}$.
According to the quotient rule,
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}$
By applying this rule to given function, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \dfrac{1-{{x}^{4}}}{1+{{x}^{4}}} \right]$

& \dfrac{dy}{dx}=\dfrac{(1+{{x}^{4}})\dfrac{d}{dx}(1-{{x}^{4}})-(1-{{x}^{4}})\dfrac{d}{dx}(1+{{x}^{4}})}{{{(1+{{x}^{4}})}^{2}}} \\\ & \dfrac{dy}{dx}=\dfrac{(1+{{x}^{4}})(0-4{{x}^{3}})-(1-{{x}^{4}})(0+4{{x}^{3}})}{{{(1+{{x}^{4}})}^{2}}} \\\ & \dfrac{dy}{dx}=\dfrac{(1+{{x}^{4}})(-4{{x}^{3}})-(1-{{x}^{4}})4{{x}^{3}}}{{{(1+{{x}^{4}})}^{2}}} \\\ \end{aligned}$$ By taking ‘$$-4{{x}^{3}}$$ ’ common in the numerator, we get $$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{-4x{}^{3}[(1+{{x}^{4}})+(1-{{x}^{4}})]}{{{(1+{{x}^{4}})}^{2}}} \\\ & \dfrac{dy}{dx}=\dfrac{-4{{x}^{3}}(2)}{{{(1+{{x}^{4}})}^{2}}} \\\ \end{aligned}$$ $$\dfrac{dy}{dx}=\dfrac{-8{{x}^{3}}}{{{(1+{{x}^{4}})}^{2}}}..........(i)$$ Now, we will find $$\dfrac{dx}{dy}$$ for a given function. As, $$y=\dfrac{1-{{x}^{4}}}{1+{{x}^{4}}}$$ By applying componendo and dividendo rule, we have $$\dfrac{y-1}{y+1}=\dfrac{(1-{{x}^{4}})-(1+{{x}^{4}})}{(1-{{x}^{4}})+(1+{{x}^{4}})}$$ $$\Rightarrow \dfrac{y-1}{y+1}=\dfrac{1-{{x}^{4}}-1-{{x}^{4}}}{1-{{x}^{4}}+1+{{x}^{4}}}$$ Cancelling the like terms, we have $$\begin{aligned} & \Rightarrow \dfrac{y-1}{y+1}=\dfrac{-2{{x}^{4}}}{2} \\\ & \Rightarrow \dfrac{y-1}{y+1}=-{{x}^{4}} \\\ & \Rightarrow {{x}^{4}}=\dfrac{-(y-1)}{y+1} \\\ & \Rightarrow {{x}^{4}}=\dfrac{1-y}{1+y} \\\ \end{aligned}$$ Now, by taking derivative of with respect to y, we have $$\dfrac{d({{x}^{4}})}{dy}=\dfrac{d}{dy}\left[ \dfrac{1-y}{1+y} \right]$$ Again, by applying the quotient rule, we have $$4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{(1+y)\dfrac{d}{dy}(1-y)-(1-y)\dfrac{d}{dy}(1+y)}{{{(1+y)}^{2}}}$$ $$\begin{aligned} & 4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{(1+y)(0-1)-(1-y)(0+1)}{{{(1+y)}^{2}}} \\\ & \Rightarrow 4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{(1+y)(-1)-(1-y)(1)}{{{(1+y)}^{2}}} \\\ & \Rightarrow 4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{-1-y-1+y}{{{(1+y)}^{2}}} \\\ \end{aligned}$$ $$4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{-2}{{{(1+y)}^{2}}}$$ Dividing throughout by ‘2’, we get $$\Rightarrow \dfrac{dx}{dy}=\dfrac{-1}{2{{x}^{3}}{{(1+y)}^{2}}}.........(ii)$$ Now as we have $$y=\dfrac{1-{{x}^{4}}}{1+{{x}^{4}}}$$. Adding ‘1’ on both sides, we get $$\begin{aligned} & 1+y=1+\dfrac{1-{{x}^{4}}}{1+{{x}^{4}}} \\\ & 1+y=\dfrac{(1+{{x}^{4}})+(1-{{x}^{4}})}{1+{{x}^{4}}} \\\ & 1+y=\dfrac{2}{1+{{x}^{4}}}.........(iii) \\\ \end{aligned}$$ Substituting equation (iii) in equation (ii), we get $$\begin{aligned} & \dfrac{dx}{dy}=\dfrac{-1}{2{{x}^{3}}{{\left( \dfrac{2}{1+{{x}^{4}}} \right)}^{2}}} \\\ & \Rightarrow \dfrac{dx}{dy}=\dfrac{-{{(1+{{x}^{4}})}^{2}}}{2{{x}^{3}}{{(2)}^{2}}} \\\ & \dfrac{dx}{dy}=\dfrac{-{{(1+{{x}^{4}})}^{2}}}{8{{x}^{3}}}.........(iv) \\\ \end{aligned}$$ Now multiplying equation (i) and (iv), we get $$\dfrac{dy}{dx}.\dfrac{dx}{dy}=\dfrac{-8{{x}^{3}}}{{{(1+{{x}^{4}})}^{2}}}.\dfrac{-{{(1+{{x}^{4}})}^{2}}}{8{{x}^{3}}}$$ Cancelling the like terms, we get $$\dfrac{dy}{dx}.\dfrac{dx}{dy}=1$$ Therefore, the correct answer is option (a). Answer is option (a) Note: In this problem we can also directly get the answer by cancelling the like terms, i.e., $$\dfrac{dy}{dx}.\dfrac{dx}{dy}=1$$