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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Suganya jagath
Given, sinu=x8+y8+z8x+2y+3z Differentiating on both sides, we get, cosu∂x∂u=∂x∂(x8+y8+z8x+2y+3z) ⇒∂x∂u=sinutanu∂x∂(x8+y8+z8x+2y+3z) ⇒∑x∂x∂u=x+2y+3zx8+y8+z8tanu((x8+y8+z8)x8+y8+z8(x8+y8+z8)(1x+2y+3z)−(x+2y+3z)(4x7x+4y7y+4z7z)) ⇒∑x∂x∂u=tanu(1−4) ⇒∑x∂x∂u=−3tanu hope you understand