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EXERCISE 1.3 · Q145

Q.If sin⁡−1x5−y5x5+y5=π6\sin^{-1}\dfrac{x^5-y^5}{x^5+y^5}=\dfrac{\pi}{6}, show dydx=x43y4\dfrac{dy}{dx}=\dfrac{x^4}{3y^4}.

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Given sin⁡−1x5−y5x5+y5=π6\sin^{-1}\dfrac{x^5-y^5}{x^5+y^5}=\dfrac{\pi}{6}.

Step 1 — undo sin⁡−1\sin^{-1}: x5−y5x5+y5=sin⁡π6=12\dfrac{x^5-y^5}{x^5+y^5}=\sin\dfrac{\pi}{6}=\dfrac12.

Step 2 — cross-multiply:

2(x5−y5)=x5+y52(x^5-y^5)=x^5+y^5

2x5−2y5=x5+y5  ⟹  x5=3y52x^5-2y^5=x^5+y^5 \implies x^5=3y^5

Step 3 — differentiate x5=3y5x^5=3y^5 implicitly: …

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