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
Verified
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 \[\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \dfrac{1-{{x}^{4}}}{1+{{x}^{4}}} \right]\] \[\begin{align} & \dfrac{dy}{dx}=\dfrac{(1+{{x}^{4}})\dfrac{d}{dx}(1-{{x}^{4}})-(1-{{x}^{4}})\dfrac{d}{dx}(1+{{x}^{4}})}{{{(1+{{x}^{4}})}^{2}}} \\ & \dfrac{dy}{dx}=\dfrac{(1+{{x}^{4}})(0-4{{x}^{3}})-(1-{{x}^{4}})(0+4{{x}^{3}})}{{{(1+{{x}^{4}})}^{2}}} \\ & \dfrac{dy}{dx}=\dfrac{(1+{{x}^{4}})(-4{{x}^{3}})-(1-{{x}^{4}})4{{x}^{3}}}{{{(1+{{x}^{4}})}^{2}}} \\ \end{align}\] By taking ‘\[-4{{x}^{3}}\] ’ common in the numerator, we get \[\begin{align} & \dfrac{dy}{dx}=\dfrac{-4x{}^{3}[(1+{{x}^{4}})+(1-{{x}^{4}})]}{{{(1+{{x}^{4}})}^{2}}} \\ & \dfrac{dy}{dx}=\dfrac{-4{{x}^{3}}(2)}{{{(1+{{x}^{4}})}^{2}}} \\ \end{align}\] \[\dfrac{dy}{dx}=\dfrac{-8{{x}^{3}}}{{{(1+{{x}^{4}})}^{2}}}..........(i)\] 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 \[\begin{align} & \Rightarrow \dfrac{y-1}{y+1}=\dfrac{-2{{x}^{4}}}{2} \\ & \Rightarrow \dfrac{y-1}{y+1}=-{{x}^{4}} \\ & \Rightarrow {{x}^{4}}=\dfrac{-(y-1)}{y+1} \\ & \Rightarrow {{x}^{4}}=\dfrac{1-y}{1+y} \\ \end{align}\] 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 \[4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{(1+y)\dfrac{d}{dy}(1-y)-(1-y)\dfrac{d}{dy}(1+y)}{{{(1+y)}^{2}}}\] \[\begin{align} & 4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{(1+y)(0-1)-(1-y)(0+1)}{{{(1+y)}^{2}}} \\ & \Rightarrow 4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{(1+y)(-1)-(1-y)(1)}{{{(1+y)}^{2}}} \\ & \Rightarrow 4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{-1-y-1+y}{{{(1+y)}^{2}}} \\ \end{align}\] \[4{{x}^{3}}\dfrac{dx}{dy}=\dfrac{-2}{{{(1+y)}^{2}}}\] 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 \[\begin{align} & 1+y=1+\dfrac{1-{{x}^{4}}}{1+{{x}^{4}}} \\ & 1+y=\dfrac{(1+{{x}^{4}})+(1-{{x}^{4}})}{1+{{x}^{4}}} \\ & 1+y=\dfrac{2}{1+{{x}^{4}}}.........(iii) \\ \end{align}\] Substituting equation (iii) in equation (ii), we get \[\begin{align} & \dfrac{dx}{dy}=\dfrac{-1}{2{{x}^{3}}{{\left( \dfrac{2}{1+{{x}^{4}}} \right)}^{2}}} \\ & \Rightarrow \dfrac{dx}{dy}=\dfrac{-{{(1+{{x}^{4}})}^{2}}}{2{{x}^{3}}{{(2)}^{2}}} \\ & \dfrac{dx}{dy}=\dfrac{-{{(1+{{x}^{4}})}^{2}}}{8{{x}^{3}}}.........(iv) \\ \end{align}\] 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., \[\dfrac{dy}{dx}.\dfrac{dx}{dy}=1\]
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