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liq [111]
3 years ago
6

Derivative of sin(x+y)

Mathematics
1 answer:
vlabodo [156]3 years ago
4 0

Answer:

\frac{dy}{dx}  =  \frac{ \cos(x + y)}{[1 - \cos(x + y)] }

Step-by-step explanation:

Let y = sin(x+y)

Differentiating with respect to x on both sides.

\frac{dy}{dx}  =  \frac{d}{dx}  \sin(x + y) \\  \\  \frac{dy}{dx}  =  \cos(x + y)\frac{d}{dx}  (x + y) \\  \\ \frac{dy}{dx}  =  \cos(x + y)(1 + \frac{dy}{dx}) \\  \\  \frac{dy}{dx}  =  \cos(x + y) +\cos(x + y) \frac{dy}{dx} \\  \\  \frac{dy}{dx}   - \cos(x + y) \frac{dy}{dx} =  \cos(x + y) \\  \\  \frac{dy}{dx}  [1 - \cos(x + y)] =  \cos(x + y) \\  \\ \purple {\bold {\frac{dy}{dx}  =  \frac{ \cos(x + y)}{[1 - \cos(x + y)] }}}

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