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`        what is the diff. of      sin6x+cos6x/sin3x+cos3x`
8 years ago

ashish kumar
17 Points
```
Dear Mustafa,
Its diferentiation will be : -
( 6cos6x  - 6sin6x ) /(3cos3x - 3sin3x )
or, 2 ( cos6x - sin6x ) / ( cos3x - cos3x )
Thank you.
```
8 years ago
ashish kumar
17 Points
```

Dear Mustafa,

Its diferentiation will be : -

( 6cos6x  - 6sin6x ) /(3cos3x - 3sin3x )

or, 2 ( cos6x - sin6x ) / ( cos3x - cos3x )

Thank you.

```
8 years ago
147 Points
```										Hi mustafa
y=(sin6x+cos6x)/(sin3x+cos3x)
dy/dx ={(sin3x+cos3x)(6cos6x-6sin6x)-(3cos3x-3sin3x)(sin6x+cos6x)}/(sin3x+cos3x)2

Now simplyfy
dy/dx =(3 cos9x -9sin3x)/ (sin3x+cos3x)2

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```
8 years ago
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