Calculate the derivative of the function

f(x)=(sin x + cos x)/(2sin x - 3 cosx).

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istetz,

in the chain rule, the derivative of the function f(x)/g(x) is

{f'(x)*g(x) - f(x)*g'(x)}/[{g(x)}^2]

So, if we derive f(x) = (sinx + cosx)/(2sinx + 3cosx),

f'(x) =

{(cosx-sinx)*(2sinx-3cosx)-(sinx+cosx)*(2cosx+3sinx)}/{(2sinx - 3sinx)^2}

= -3(cosx)^2 - 2(sinx)^2 - 3(sinx)^2 -2(cosx)^2

Since (sinx)^2 + (cosx)^2 = 1,

the value of the equation above is

-3 -2 = -5

Hence, the derivative of f(x) is -5

You need to use the quotient rule to find the derivative of `f(x)=(sin x + cos x)/(2sin x - 3 cosx).`

Differentiating with respect to x yields:

`f'(x) = ((sin x + cos x)'*(2sin x - 3 cosx) - (sin x + cos x)*(2sin x - 3 cosx)')/(2sin x - 3 cosx)^2`

`f'(x) = ((cos x - sin x)*(2sin x - 3 cosx) - (sin x + cos x)*(2cos x + 3 sin x))/(2sin x - 3 cosx)^2`

Opening the brackets yields:

`f'(x) = (2sin x*cos x - 3cos^2 x - 2sin^2 x + 3 sin x*cos x - 2 sin x*cos x+ 3 sin^2 x + 2 cos^2 x + 3 sin x*cos x)/(2sin x - 3 cosx)^2`

`` `f'(x) = (6 sin x*cos x+ sin^2 x - cos^2 x)/(2sin x - 3 cosx)^2`

`f'(x) = (3 sin 2x- cos 2x)/(2sin x - 3 cosx)^2`

**The derivative of the given function: `f'(x) = (3 sin 2x- cos 2x)/(2sin x - 3 cosx)^2` **

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