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Question
Find `bb(dy/dx)` in the following:
ax + by2 = cos y
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Solution
ax + by2 = cos y
Differentiating both sides with respect to x,
⇒ `a d/dx (x) + b d/dx (y^2) = d/dx(cos y)`
⇒ `a xx 1 + b * 2y dy/dx = - sin y dy/dx`
⇒ `a + 2by dy/dx + sin y dy/dx = 0`
⇒ `a + dy/dx (2by +sin y) = 0`
⇒ `dy/dx (2by + sin y) = - a`
∴ `dy/dx = (-a)/(2by + sin y)`
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