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 #