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Implicit differentiation/Trig problem?
The problem is: Find the gradient of the curve 2sin(xy)=1 when y=1/2 and pi<x<2pi. Please explain as well
1 Answer
- ?Lv 76 years ago
When y = 1/2 we have:
2sin(x/2) = 1
=> sin(x/2) = 1/2
so, x/2 = π/6 + 2nπ or 5π/6 + 2nπ...for n ∈ ℕ
=> x = π/3 + 4nπ or 5π/3 + 4nπ
Therefore, as π < x < 2π we take x = 5π/3
Now, differentiating implicitly we get:
2[xcos(xy)(dy/dx) + ycos(xy)] = 0
=> dy/dx = -[ycos(xy)]/[xcos(xy)]
i.e. dy/dx = -y/x
Hence, dy/dx = -(1/2)/(5π/3) => -3/(10π)
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