Question

If y = a sin x + b cos x, find dy/dx when tan x =a/b

Answer 1

Answer:

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Answer 2

$tanx = \frac{a}{b} \\ \\ {sec}^{2} x = 1 + {tan}^{2} x \\ secx = \sqrt{1 + \frac{ {a}^{2} }{ {b}^{2} } } \\ secx = \frac{ \sqrt{ {a}^{2} + {b}^{2} } }{b} \\ \\ cosx = \frac{b}{ \sqrt{ {a}^{2} + {b}^{2} } } \\ \\ sinx = \sqrt{1 - \frac{ {b}^{2} }{ {a}^{2} + {b}^{2} } } \\ sinx = \frac{a}{ \sqrt{ {a}^{2} + {b}^{2} } } \\ \\ \\ y = asinx + bcosx \\ y = a \times \frac{a}{ \sqrt{ {a}^{2} + {b}^{2} } } + b \times \frac{b}{ \sqrt{ {a}^{2} + {b}^{2} } } \\ y = \frac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } + \frac{ {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } } \\ \\ \frac{dy}{dx} = 0$