For the curve C:(x²+y²-3)+(x²-y²-1)⁵=0, the value of 3 y′-y³ y^(′ ′), at the point (α, α), α>0, on C, is equal to ____________.

Solution

Maths sir · 4 steps

Call u = x² - y² - 1. At (α, α) it equals -1, and that makes every derivative easy.

1. Find the point. At (α, α) we get u = α² - α² - 1 = -1. So the curve gives 2α² - 3 + (-1)⁵ = 0, which means α² = 2 and α = √2. At this point u⁴ = 1 and u³ = -1.
2. First derivative. Differentiate and divide by 2: x + yy' + 5u⁴(x - yy') = 0. At x = y = √2 this is √2(1 + y') + 5√2(1 - y') = 0, so 6 - 4y' = 0 and y' = 3/2.

   1 + y'² + yy'' + 40u³(x - yy')² + 5u⁴(1 - y'² - yy'') = 0

3. Second derivative. This comes from differentiating the first-derivative equation once more, using u' = 2(x - yy'). At the point, x - yy' = √2 - 3√2/2 = -√2/2, so (x - yy')² = 1/2. Then 1 + 9/4 + yy'' - 20 + 5 - 45/4 - 5yy'' = 0, which gives -23 - 4yy'' = 0 and yy'' = -23/4.
4. So. y³ y'' = y² · yy'' = 2 × (-23/4) = -23/2. Then 3y' - y³ y'' = 9/2 + 23/2 = 16.

The common slip: 6. Treating u⁴ as the constant 1 before the second differentiation drops the 40u³(x - yy')² term, which gives yy'' = -3/4 and a final value of 6.

The idea to keep. In implicit differentiation, substitute the point as soon as you have each derivative. Giving a messy bracket a name like u keeps the chain rule terms visible, so none of them gets dropped.

Answer from the NTA final answer key. Solution by PrepWiser.