Question

If $\left| t \right|<1,\sin x=\dfrac{2t}{1+{{t}^{2}}},\tan y=\dfrac{2t}{1-{{t}^{2}}}$, then
$\dfrac{dy}{dx}$ is equal to
(a) $\dfrac{1}{x}$
(b) $\dfrac{1}{2}$
(c) $\dfrac{-1}{2}$
(d) $\dfrac{-1}{x}$
(e) $1$

Answer 1

To find the value of $\dfrac{dy}{dx}$, first evaluate $\dfrac{dy}{dt}$and $\dfrac{dx}{dt}$. Then use the formula $\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{1}{\dfrac{dx}{dt}}$ to get the value of $\dfrac{dy}{dx}$.

Complete step-by-step answer:
We have the parametric equations $\left| t \right|<1,\sin x=\dfrac{2t}{1+{{t}^{2}}},\tan y=\dfrac{2t}{1-{{t}^{2}}}$. We have to evaluate $\dfrac{dy}{dx}$.
We will first evaluate the values of $\dfrac{dy}{dt}$and $\dfrac{dx}{dt}$. Then, we will rearrange the terms such that $\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{1}{\dfrac{dx}{dt}}$to get the value of $\dfrac{dy}{dx}$.
We will begin by finding the value of $\dfrac{dx}{dt}$. We have the parametric equation $\sin x=\dfrac{2t}{1+{{t}^{2}}}$. We will differentiate the given equation on both sides with respect to the variable $t$. Thus, we have $\dfrac{d}{dt}\left( \sin x \right)=\dfrac{d}{dt}\left( \dfrac{2t}{1+{{t}^{2}}} \right).....\left( 1 \right)$.
To find the value of $\dfrac{d}{dt}\left( \sin x \right)$, we will multiply and divide the equation by $dx$. Thus, we have $\dfrac{d}{dt}\left( \sin x \right)=\dfrac{d}{dx}\left( \sin x \right)\times \dfrac{dx}{dt}$.
We know that the first derivative of the function $y=\sin x$ is $\dfrac{dy}{dx}=\cos x$.
Thus, we have $\dfrac{d}{dt}\left( \sin x \right)=\dfrac{d}{dx}\left( \sin x \right)\times \dfrac{dx}{dt}=\cos x\dfrac{dx}{dt}$.
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. So, we can write $\cos x=\sqrt{1-{{\sin }^{2}}x}$.
Substituting $\sin x=\dfrac{2t}{1+{{t}^{2}}}$ in the above equation, we get $\cos x=\sqrt{1-{{\sin }^{2}}x}=\sqrt{1-{{\left( \dfrac{2t}{1+{{t}^{2}}} \right)}^{2}}}=\dfrac{\sqrt{{{\left( 1+{{t}^{2}} \right)}^{2}}-4{{t}^{2}}}}{1+{{t}^{2}}}=\dfrac{\sqrt{1+{{t}^{4}}+2{{t}^{2}}-4{{t}^{2}}}}{1+{{t}^{2}}}=\dfrac{\sqrt{1+{{t}^{4}}-2{{t}^{2}}}}{1+{{t}^{2}}}=\dfrac{\sqrt{{{\left( 1-{{t}^{2}} \right)}^{2}}}}{1+{{t}^{2}}}=\dfrac{1-{{t}^{2}}}{1+{{t}^{2}}}$
Hence, we have $\dfrac{d}{dt}\left( \sin x \right)=\cos x\dfrac{dx}{dt}=\dfrac{1-{{t}^{2}}}{1+{{t}^{2}}}\dfrac{dx}{dt}.....\left( 2 \right)$.
Now, we will evaluate the value of $\dfrac{d}{dt}\left( \dfrac{2t}{1+{{t}^{2}}} \right)$. To do so, we will use quotient rule of differentiation which states that if $y=\dfrac{f\left( x \right)}{g\left( x \right)}$ then we have $\dfrac{dy}{dx}=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x \right)g'\left( x \right)}{{{g}^{2}}\left( x \right)}$.
Thus, we have $\dfrac{d}{dt}\left( \dfrac{2t}{1+{{t}^{2}}} \right)=\dfrac{\left( 1+{{t}^{2}} \right)\dfrac{d}{dt}\left( 2t \right)-2t\dfrac{d}{dt}\left( 1+{{t}^{2}} \right)}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$.
We know that derivative of any function of the form $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have $\dfrac{d}{dt}\left( \dfrac{2t}{1+{{t}^{2}}} \right)=\dfrac{\left( 1+{{t}^{2}} \right)\dfrac{d}{dt}\left( 2t \right)-2t\dfrac{d}{dt}\left( 1+{{t}^{2}} \right)}{{{\left( 1+{{t}^{2}} \right)}^{2}}}=\dfrac{\left( 1+{{t}^{2}} \right)\left( 2 \right)-\left( 2t \right)\left( 2t \right)}{{{\left( 1+{{t}^{2}} \right)}^{2}}}=\dfrac{2t+2{{t}^{2}}-4{{t}^{2}}}{{{\left( 1+{{t}^{2}} \right)}^{2}}}=\dfrac{2\left( 1-{{t}^{2}} \right)}{{{\left( 1+{{t}^{2}} \right)}^{2}}}.....\left( 3 \right)$. Substituting equation $\left( 2 \right),\left( 3 \right)$ in equation $\left( 1 \right)$, we get $\dfrac{1-{{t}^{2}}}{1+{{t}^{2}}}\dfrac{dx}{dt}=\dfrac{2\left( 1-{{t}^{2}} \right)}{{{\left( 1+{{t}^{2}} \right)}^{2}}}$.
Simplifying the above equation, we have $\dfrac{dx}{dt}=\dfrac{2\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)\left( 1-{{t}^{2}} \right)}=\dfrac{2}{\left( 1+{{t}^{2}} \right)}.....\left( 4 \right)$.
We will now find the value of $\dfrac{dy}{dt}$. We have the parametric equation $\tan y=\dfrac{2t}{1-{{t}^{2}}}$. We will differentiate the given equation on both sides with respect to the variable $t$. Thus, we have $\dfrac{d}{dt}\left( \tan y \right)=\dfrac{d}{dt}\left( \dfrac{2t}{1-{{t}^{2}}} \right).....\left( 5 \right)$.
To find the value of $\dfrac{d}{dt}\left( \tan y \right)$, we will multiply and divide the equation by $dy$. Thus, we have $\dfrac{d}{dt}\left( \tan y \right)=\dfrac{d}{dy}\left( \tan y \right)\times \dfrac{dy}{dt}$.
We know that the first derivative of the function $y=\tan x$ is $\dfrac{dy}{dx}={{\sec }^{2}}x$.
Thus, we have $\dfrac{d}{dt}\left( \tan y \right)=\dfrac{d}{dy}\left( \tan y \right)\times \dfrac{dy}{dt}={{\sec }^{2}}y\dfrac{dy}{dt}$.
We know that $1+{{\tan }^{2}}y={{\sec }^{2}}y$.
Substituting $\tan y=\dfrac{2t}{1-{{t}^{2}}}$ in the above equation, we get ${{\sec }^{2}}y=1+{{\tan }^{2}}y=1+{{\left( \dfrac{2t}{1-{{t}^{2}}} \right)}^{2}}=\dfrac{{{\left( 1-{{t}^{2}} \right)}^{2}}+4{{t}^{2}}}{{{\left( 1-{{t}^{2}} \right)}^{2}}}=\dfrac{1+{{t}^{4}}-2{{t}^{2}}+4{{t}^{2}}}{{{\left( 1-{{t}^{2}} \right)}^{2}}}=\dfrac{1+{{t}^{4}}+2{{t}^{2}}}{{{\left( 1-{{t}^{2}} \right)}^{2}}}=\dfrac{{{\left( 1+{{t}^{2}} \right)}^{2}}}{{{\left( 1-{{t}^{2}} \right)}^{2}}}$
Hence, we have $\dfrac{d}{dt}\left( \tan y \right)={{\sec }^{2}}y\dfrac{dy}{dt}=\dfrac{{{\left( 1+{{t}^{2}} \right)}^{2}}}{{{\left( 1-{{t}^{2}} \right)}^{2}}}\dfrac{dy}{dt}.....\left( 6 \right)$.
Now, we will evaluate the value of $\dfrac{d}{dt}\left( \dfrac{2t}{1-{{t}^{2}}} \right)$. To do so, we will use quotient rule of differentiation which states that if $y=\dfrac{f\left( x \right)}{g\left( x \right)}$ then we have $\dfrac{dy}{dx}=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x \right)g'\left( x \right)}{{{g}^{2}}\left( x \right)}$.
Thus, we have $\dfrac{d}{dt}\left( \dfrac{2t}{1-{{t}^{2}}} \right)=\dfrac{\left( 1-{{t}^{2}} \right)\dfrac{d}{dt}\left( 2t \right)-2t\dfrac{d}{dt}\left( 1-{{t}^{2}} \right)}{{{\left( 1-{{t}^{2}} \right)}^{2}}}$.
We know that derivative of any function of the form $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have $\dfrac{d}{dt}\left( \dfrac{2t}{1-{{t}^{2}}} \right)=\dfrac{\left( 1-{{t}^{2}} \right)\dfrac{d}{dt}\left( 2t \right)-2t\dfrac{d}{dt}\left( 1-{{t}^{2}} \right)}{{{\left( 1-{{t}^{2}} \right)}^{2}}}=\dfrac{\left( 1-{{t}^{2}} \right)\left( 2 \right)-\left( 2t \right)\left( -2t \right)}{{{\left( 1-{{t}^{2}} \right)}^{2}}}=\dfrac{2-2{{t}^{2}}+4{{t}^{2}}}{{{\left( 1-{{t}^{2}} \right)}^{2}}}=\dfrac{2\left( 1+{{t}^{2}} \right)}{{{\left( 1-{{t}^{2}} \right)}^{2}}}.....\left( 7 \right)$. Substituting equation $\left( 6 \right),\left( 7 \right)$in equation $\left( 5 \right)$, we get $\dfrac{{{\left( 1+{{t}^{2}} \right)}^{2}}}{{{\left( 1-{{t}^{2}} \right)}^{2}}}\dfrac{dy}{dt}=\dfrac{2\left( 1+{{t}^{2}} \right)}{{{\left( 1-{{t}^{2}} \right)}^{2}}}$.
Simplifying the above equation, we have $\dfrac{dy}{dt}=\dfrac{2}{\left( 1+{{t}^{2}} \right)}.....\left( 8 \right)$.
Dividing equation $\left( 8 \right)$ by equation $\left( 4 \right)$, we get $\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{2}{\left( 1+{{t}^{2}} \right)}}{\dfrac{2}{\left( 1+{{t}^{2}} \right)}}$.
Simplifying the above expression, we have $\dfrac{dy}{dx}=1$.
Hence, given $\left| t \right|<1,\sin x=\dfrac{2t}{1+{{t}^{2}}},\tan y=\dfrac{2t}{1-{{t}^{2}}}$, the value of $\dfrac{dy}{dx}$is $1$, which is option (e).

Note: We can also solve this question by simplifying the expression using inverse trigonometric functions. For $\left| t \right|<1$, we can write $x={{\sin }^{-1}}\left( \dfrac{2t}{1+{{t}^{2}}} \right)=2{{\tan }^{-1}}t,y={{\tan }^{-1}}\left( \dfrac{2t}{1-{{t}^{2}}} \right)=2{{\tan }^{-1}}t$ and thus, we can write $x=y$ and differentiate it to find the derivative.