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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
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148.2k+ views
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.