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Question: If \({{x}^{2}}+{{y}^{2}}=t-\dfrac{1}{t}\), \({{x}^{4}}+{{y}^{4}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}\), t...

If x2+y2=t−1t{{x}^{2}}+{{y}^{2}}=t-\dfrac{1}{t}, x4+y4=t2+1t2{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}, then x3ydydx{{x}^{3}}y\dfrac{dy}{dx} is equal to
a. 1
b. 2
c. 3
d. 4

Explanation

Solution

To obtain the value of a given differential equation we will use implicit differentiation. Firstly by using the two values given we will find a relation between xx and yy. Then using implicit differentiation we will differentiate the value obtained. Finally we will keep the terms whose value we have to find on one side and the rest on the other side and get the desired answer.

Complete step by step solution:
It is given to us that:
x2+y2=t−1t{{x}^{2}}+{{y}^{2}}=t-\dfrac{1}{t}……(1)\left( 1 \right)
x4+y4=t2+1t2{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}……..(2)\left( 2 \right)
Now we will take square of the equation (1) and use the formula (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab as follows:
x2+y2=t−1t ⇒(x2+y2)2=(t−1t)2 ⇒x4+y4+2x2y2=t2+1t2−2 \begin{aligned} & {{x}^{2}}+{{y}^{2}}=t-\dfrac{1}{t} \\\ & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( t-\dfrac{1}{t} \right)}^{2}} \\\ & \Rightarrow {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2 \\\ \end{aligned}
Now we will replace value from equation (2) above we get:
x4+y4+2x2y2=x4+y4−2 ⇒2x2y2=x4+y4−x4−y4−2 ⇒2x2y2=−2 ⇒y2=−x−2 \begin{aligned} & {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{x}^{4}}+{{y}^{4}}-2 \\\ & \Rightarrow 2{{x}^{2}}{{y}^{2}}={{x}^{4}}+{{y}^{4}}-{{x}^{4}}-{{y}^{4}}-2 \\\ & \Rightarrow 2{{x}^{2}}{{y}^{2}}=-2 \\\ & \Rightarrow {{y}^{2}}=-{{x}^{-2}} \\\ \end{aligned}
Next we will use Implicit differentiation and differentiate above value with respect to xx as follows:
y2=−x−2 ⇒2ydydx=−(−2)x−3 ⇒ydydx=x−3 \begin{aligned} & {{y}^{2}}=-{{x}^{-2}} \\\ & \Rightarrow 2y\dfrac{dy}{dx}=-\left( -2 \right){{x}^{-3}} \\\ & \Rightarrow y\dfrac{dy}{dx}={{x}^{-3}} \\\ \end{aligned}
Now as we need the value of x3ydydx{{x}^{3}}y\dfrac{dy}{dx} we will rewrite the above value as:
1x−3ydydx=1 ∴x3ydydx=1 \begin{aligned} & \dfrac{1}{{{x}^{-3}}}y\dfrac{dy}{dx}=1 \\\ & \therefore {{x}^{3}}y\dfrac{dy}{dx}=1 \\\ \end{aligned}

Hence x3ydydx{{x}^{3}}y\dfrac{dy}{dx} is equal to 1.

Note:
Differentiation is a method to find the derivative of the given function. If two unknown variables are given then the rate change of one with respect to the other is known as the derivatives. Implicit differentiation is used when on differentiating one term with respect to another we can’t lead to the answer directly. In implicit differentiation we get the derivative value as well which has to be taken together in order to get the answer. The algebraic formula used is the square of the sum of two numbers. We can simply multiply the two terms also but that will make our calculation lengthy.