How do you use the quotient rule to find the derivative of #y=(1+cos(x))/(1+sin(x))# ?

1 Answer

The quotient rule states that given functions u and v such that #y = (u(x))/(v(x)), dy/dx = (u'(x)v(x) - u(x)v'(x))/(v^2(x))# By assigning u and v equal to the numerator and denominator, respectively, in the example function, we arrive at #dy/dx = [(-sin x)(1+sin x) - (1+cos x)(cos x)]/(1+sin x)^2#

When using the Quotient Rule, it is important to designate your functions #u(x)# and #v(x)# in such a way as to make things simple while still being accurate. In the above case, declaring #cos(x)# as #u(x)# would not allow us to use the quotient rule efficiently, as our numerator would then be #1 + u(x)#.

From the identities of trigonometric function derivatives, we know that the derivative of #sin(x)# is #cos(x)#, and the derivative of #cos(x)# is #-sin(x)#. This could be proven using Euler's Formula, but for our purposes we shall accept these without proof. The derivative of any constant (such as 1 in our example) is 0, and the derivative of a sum is equal to the sum of the derivatives. Therefore:

#u(x) = 1+cos(x), u'(x) = -sin(x), v(x) = 1+sin(x), v'(x) = cos(x)#

#dy/dx = [(-sin x)(1+sin x) - (1+cos x)(cos x)]/(1+sin x)^2 #

Attempting to use trig identities to simplify...

#= [-sin(x) - sin^2(x) - cos(x) - cos^2(x)]/(1+sin x)^2 = [-1(sin^2(x)+cos^2(x)) -1 (sin x + cos x)]/(1+sin x)^2 = -[1 + sin(x) + cos(x)]/(1+sin(x))^2#

However, the initial answer should be sufficient.