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If y=(sin(x+9))/(cosx), then (dy)/(dx) a...

If `y=(sin(x+9))/(cosx)`, then `(dy)/(dx)` at x = 0 is: a. `cos9` b. `sin9` c. 0 d. 1

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Given `y=(sin(x+9))/(cosx)`
Differentiating we get
By quotient formula
`(dy/dx)​=(cosx(d/dx)(sin(x+9))−sin(x+9)(d/dx)​cosx)/cos^2x`
​ ⇒`(dy/dx)​=(cosxcos(x+9)−sin(x+9)(−sinx))/cos^2x`
​ ...
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