Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If x2+y2=t−1tx^2 + y^2 = t - \dfrac{1}{t} and x4+y4=t2+1t2x^4 + y^4 = t^2 + \dfrac{1}{t^2}, then x3ydydxx^3 y\dfrac{dy}{dx} is.
Solution
Answer: 1
Step 1: Relate the two equations
(x2+y2)2=(t−1t)2=t2−2+1t2\left(x^2+y^2\right)^2 = \left(t-\frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2}
x4+y4+2x2y2=t2−2+1t2x^4+y^4+2x^2y^2 = t^2 - 2 + \frac{1}{t^2}
 Substituting x4+y4=t2+1t2x^4+y^4 = t^2+\dfrac{1}{t^2}:
t2+1t2+2x2y2=t2−2+1t2  ⟹  x2y2=−1t^2+\frac{1}{t^2}+2x^2y^2 = t^2-2+\frac{1}{t^2} \implies x^2y^2 = -1
 Step 2: Differentiate and extract the required expression Differentiating x2y2=−1x^2y^2 = -1 with respect to xx:
2xy2+2x2ydydx=0  ⟹  dydx=−yx2xy^2 + 2x^2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}
x3ydydx=x3y⋅(−yx)=−x2y2=−(−1)=1x^3y\frac{dy}{dx} = x^3y\cdot\left(-\frac{y}{x}\right) = -x^2y^2 = -(-1) = 1
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