Question

If $\tan y = \dfrac{{2t}}{{1 - {t^2}}}$ and $\sin x = \dfrac{{2t}}{{1 + {t^2}}}$, then $\dfrac{{dy}}{{dx}}$ equals:
(A) $\dfrac{2}{{1 - {t^2}}}$
(B) $\dfrac{1}{{1 + {t^2}}}$
(C) $1$
(D) $2$

Answer 1

In the given problem, we are required to differentiate a parametric function in parameter t.
For differentiating parametric functions, we first find derivatives of x and y with respect to t separately and then divide both of them to find differential $\dfrac{{dy}}{{dx}}$. Also, we will use the chain rule of differentiation in order to solve the problem. We must remember the derivatives of trigonometric functions such as sine and cosine to get to the final answer.

Complete answer:
Now, we have, $\tan y = \dfrac{{2t}}{{1 - {t^2}}}$ and $\sin x = \dfrac{{2t}}{{1 + {t^2}}}$. So, to find the derivative of the parametric function, we first differentiate the functions in x and y separately with respect to the parameter, t.
Hence, differentiating the functions with respect to t, we get,
$\dfrac{d}{{dt}}\left( {\tan y} \right) = \dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 - {t^2}}}} \right)$
We know that the derivative of tangent function $\tan x$ with respect to x is ${\sec ^2}x$. So, we get,
$\Rightarrow {\sec ^2}y \times \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 - {t^2}}}} \right)$
Now, we find derivative of right side of equation using quotient rule of differentiation $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$. So, we get,
$\Rightarrow {\sec ^2}y \times \dfrac{{dy}}{{dt}} = \dfrac{{\left( {1 - {t^2}} \right)\dfrac{d}{{dt}}\left( {2t} \right) - \left( {2t} \right)\dfrac{d}{{dt}}\left( {1 - {t^2}} \right)}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Using power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$,
$\Rightarrow {\sec ^2}y \times \dfrac{{dy}}{{dt}} = \dfrac{{\left( {1 - {t^2}} \right)\left( 2 \right) - \left( {2t} \right)\left( { - 2t} \right)}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
$\Rightarrow {\sec ^2}y \times \dfrac{{dy}}{{dt}} = \dfrac{{2 - 2{t^2} + 4{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Opening the brackets,
$\Rightarrow {\sec ^2}y \times \dfrac{{dy}}{{dt}} = \dfrac{{2 + 2{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
So, shifting secant term to right side of equation,
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{1}{{{{\sec }^2}y}} \times \dfrac{{2 + 2{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Using trigonometric identity ${\sec ^2}x = {\tan ^2}x + 1$, we get ,
$\Rightarrow \dfrac{{dy}}{{dt}} = \left( {\dfrac{1}{{{{\tan }^2}y + 1}}} \right) \times \dfrac{{2 + 2{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Substituting the value of tangent as $\tan y = \dfrac{{2t}}{{1 - {t^2}}}$
$\Rightarrow \dfrac{{dy}}{{dt}} = \left( {\dfrac{1}{{{{\left( {\dfrac{{2t}}{{1 - {t^2}}}} \right)}^2} + 1}}} \right) \times \dfrac{{2 + 2{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Computing the squares and taking LCM, we get,
$\Rightarrow \dfrac{{dy}}{{dt}} = \left( {\dfrac{1}{{\dfrac{{4{t^2} + {t^4} - 2{t^2} + 1}}{{{t^4} - 2{t^2} + 1}}}}} \right) \times \dfrac{{2 + 2{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Simplifying the expression,
$\Rightarrow \dfrac{{dy}}{{dt}} = \left( {\dfrac{{{t^4} - 2{t^2} + 1}}{{{t^4} + 2{t^2} + 1}}} \right) \times \dfrac{{2 + 2{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Condensing the expressions as whole squares,
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{{{\left( {1 - {t^2}} \right)}^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}} \times \dfrac{{2 + 2{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}}$
Cancelling the common terms in numerator and denominator, we get,
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{2}{{\left( {1 + {t^2}} \right)}} - - - - - \left( 1 \right)$
Now, we differentiate $\sin x = \dfrac{{2t}}{{1 + {t^2}}}$ with respect to t. So, we get,
$\dfrac{d}{{dt}}\left( {\sin x} \right) = \dfrac{d}{{dt}}\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right)$
We know that derivative of sine is cosine. Using quotient rule on right side of equation, we get,
$\Rightarrow \cos x \times \dfrac{{dx}}{{dt}} = \dfrac{{\left( {1 + {t^2}} \right)\dfrac{d}{{dt}}\left( {2t} \right) - \left( {2t} \right)\dfrac{d}{{dt}}\left( {1 + {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Using power rule of differentiation, we get,
$\Rightarrow \cos x \times \dfrac{{dx}}{{dt}} = \dfrac{{\left( {1 + {t^2}} \right)\left( 2 \right) - \left( {2t} \right)\left( {2t} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Shifting cosine to right side of equation,
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{\cos x}} \times \dfrac{{2 + 2{t^2} - 4{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{\cos x}} \times \dfrac{{2 - 2{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Using the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$, we get,
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{\sqrt {1 - {{\sin }^2}x} }} \times \dfrac{{2 - 2{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Substituting value of sine,
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{{2t}}{{1 + {t^2}}}} \right)}^2}} }} \times \dfrac{{2 - 2{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Computing whole squares and taking LCM,
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{1}{{\sqrt {\dfrac{{1 + {t^4} + 2{t^2} - 4{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}} }} \times \dfrac{{2 - 2{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Simplifying the expression, we get,
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{\left( {1 + {t^2}} \right)}}{{\left( {1 - {t^2}} \right)}} \times \dfrac{{2 - 2{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{2}{{\left( {1 + {t^2}} \right)}} - - - - - \left( 2 \right)$
Dividing equation $\left( 1 \right)$ by equation $\left( 2 \right)$, we get,
$\dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}} = \dfrac{{\left[ {\dfrac{2}{{\left( {1 + {t^2}} \right)}}} \right]}}{{\left[ {\dfrac{2}{{\left( {1 + {t^2}} \right)}}} \right]}}$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = 1$
Hence, the correct answer is the option (C).

Note:
We must remember this method to find the derivatives of the parametric function. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.
We can also solve the problem using trigonometric substitution as $t = \tan \theta$.
Then, we get, $\tan y = \dfrac{{2t}}{{1 - {t^2}}} = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
We know the double angle formula for tangent as $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$.
So, we get, $\tan y = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \tan 2\theta$.
Taking tangent inverse function on both sides of equation,
$\Rightarrow y = 2\theta$
Similarly, $\sin x = \dfrac{{2t}}{{1 + {t^2}}} = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$.
We know the double angle formula for sine in terms of tangent as $\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$. Now, we get,
$\sin x = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta$
Taking sine inverse function on both sides,
$\Rightarrow x = 2\theta$
Now, we can calculate the value of $\dfrac{{dy}}{{dx}}$ by chain rule $\dfrac{{\left( {\dfrac{{dy}}{{d\theta }}} \right)}}{{\left( {\dfrac{{dx}}{{d\theta }}} \right)}}$.
So, we get, $\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{d}{{d\theta }}\left( {2\theta } \right)}}{{\dfrac{d}{{d\theta }}\left( {2\theta } \right)}}$
Using the power rule of differentiation, we get,
$\dfrac{{dy}}{{dx}} = \dfrac{2}{2} = 1$
Hence, we get the same answer.