How do you find the derivative of $(sin x + 2x) / (cos x - 2)$ ?

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Answer 1 · 2016-01-27T17:03:55

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

Using the quotient rule, we take functions $h(x)$ and $g(x)$ as:

$g(x) = sin x + 2x$
$h(x) = cos x - 2$

Knowing that trigonometric derivatives are:

$(d(cos x))/dx = - sin x$

$(d(sin x))/dx = cos x$

We can use now the quotient rule and find the final expression for $f'(x)$ :

$f'(x) = ((cos x + 2)(cos x - 2) - (sin x + 2x)(-sin x))/(cos x -2)^2 =$

$= (cos^2x - 4+sin^2x + 2x sin x)/(cosx-2)^2$

Using the trigonometric identity:

$sin^2x+cos^2x = 1$

In the end, we have:

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

How do you find the derivative of (sin x + 2x) / (cos x - 2) (sin x + 2x) / (cos x - 2)?