Question

Find the derivative of $y$ w.r.t. $x$ where $x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}$ and $y = \dfrac{{2t}}{{1 + {t^2}}}$.
(a) $- \dfrac{x}{y}$
(b) $- \dfrac{y}{x}$
(c) 1
(d) None of the above

Answer 1

Hint : We will use the most eccentric concept of derivations. Considering the ‘ $t$ ’ variable as a derivating agent or term the solution is solved by using certain rules of derivation such as $\dfrac{d}{{dx}}{\left( x \right)^n} = n{x^{n - 1}}$ , etc. As a result, substituting the values in the equation $\dfrac{{dy}}{{dx}}$ (after finding the derivatives of each given terms i.e. $\dfrac{{dx}}{{dt}}$ and $\dfrac{{dy}}{{dt}}$ respectively) one can easily solve the complete problem.

Complete step-by-step answer :
Since, we have the given equations that
$x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}$ And $y = \dfrac{{2t}}{{1 + {t^2}}}$
As a result, solving the given equations, first of all derivating the above given equations with respect to the ‘ $t$ ’ variable, can reach up to a desire output,
Hence, derivating each term individually, we get
$x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{1 - {t^2}}}{{1 + {t^2}}}} \right)$
Derivating the above equation with respect to ‘ $t$ ’ by using laws of derivation for dividation i.e. $\dfrac{d}{{dx}}\left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \dfrac{{\left( {1 - x} \right)\dfrac{d}{{dx}}\left( {1 + x} \right) - \left( {1 + x} \right)\dfrac{d}{{dx}}\left( {1 - x} \right)}}{{{{\left( {1 - x} \right)}^2}}}$ respectively, we get,
$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{{1 - {t^2}}}{{1 + {t^2}}}} \right) = \dfrac{{\left( {1 + {t^2}} \right)\dfrac{d}{{dt}}\left( {1 - {t^2}} \right) - \left( {1 - {t^2}} \right)\dfrac{d}{{dt}}\left( {1 + {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Using the rule of derivative/s that is $\dfrac{d}{{dx}}{\left( x \right)^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0$ we get
$\dfrac{d}{{dt}}\left( {\dfrac{{1 - {t^2}}}{{1 + {t^2}}}} \right) = \dfrac{{\left( {1 + {t^2}} \right)\left( { - 2t} \right) - \left( {1 - {t^2}} \right)\left( {2t} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Solving the equation mathematically, we get

\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{{1 - {t^2}}}{{1 + {t^2}}}} \right) = \dfrac{{ - 2t\left( {1 + {t^2} + 1 - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}} \\\ \dfrac{d}{{dt}}\left( {\dfrac{{1 - {t^2}}}{{1 + {t^2}}}} \right) = \left( { - 2} \right)\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right)\left( {\dfrac{1}{{1 + {t^2}}}} \right) \\\ $$ … (i) Similarly, for $$y = \dfrac{{2t}}{{1 + {t^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right)$$ Derivating the above equation with respect to ‘ $ t $ ’ by using laws of derivation for dividation, we get, $$\dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right) = \dfrac{{\left( {1 + {t^2}} \right)\dfrac{d}{{dt}}\left( {2t} \right) - \left( {2t} \right)\dfrac{d}{{dt}}\left( {1 + {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$$ Using the rule of derivative/s that is $ \dfrac{d}{{dx}}{\left( x \right)^n} = n{x^{n - 1}} $ and $ \dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0 $ we get $$\dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right) = \dfrac{{\left( {1 + {t^2}} \right)\left( 2 \right) - \left( {2t} \right)\left( {2t} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$$ Solving the equation mathematically, we get

\dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right) = \dfrac{{2 + 2{t^2} - 4{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}} = \dfrac{{2 - 2{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}} \\
\dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right) = \dfrac{{2\left( {{1^2} - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}} \\

Now, we know that By definition of derivatives, it seems that $$\dfrac{{dy}}{{dx}} = \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$$ As a result, from (i) and (ii), Substituting all the values for the final solution i.e. $$\dfrac{{dx}}{{dt}} = \left( { - 2} \right)\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right)\left( {\dfrac{1}{{1 + {t^2}}}} \right)$$and$$\dfrac{{dy}}{{dt}} = = \dfrac{{2\left( {{1^2} - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$$, we get $$\dfrac{{dy}}{{dx}} = \dfrac{{\left[ {\dfrac{{2\left( {{1^2} - {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right]}}{{\left[ {\left( { - 2} \right)\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right)\left( {\dfrac{1}{{1 + {t^2}}}} \right)} \right]}}$$ Solving the equation mathematically, we get $$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{2\left( {1 - {t^2}} \right)}}{{\left( {1 + {t^2}} \right)}}}}{{\left( { - 2} \right)\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right)}}$$ Now, since we know that the given equation in terms of ‘ $ x $ ’ and ‘ $ y $ ’ respectively are As a result, substituting $ x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}} $ and $ y = \dfrac{{2t}}{{1 + {t^2}}} $ in the above equation, we get

\dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{\left( { - 2} \right)y}} \\
\therefore \dfrac{{dy}}{{dx}} = - \dfrac{x}{y} \\

Therefore, the option (a) is correct! **So, the correct answer is “Option a”.** **Note** : One must remember the concept of derivation, how to differentiate the equation with respect to which variable, etc.? Also, laws of derivation such as $ \dfrac{d}{{dx}}\left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \dfrac{{\left( {1 - x} \right)\dfrac{d}{{dx}}\left( {1 + x} \right) - \left( {1 + x} \right)\dfrac{d}{{dx}}\left( {1 - x} \right)}}{{{{\left( {1 - x} \right)}^2}}} $ is the rule of derivation used here. Derivation of any constant number is always zero. Deriving the equation with the same term is always one $ \dfrac{d}{{dx}}(x) = 1 $. Algebraic identities play a significant role in solving this problem.