How do you find the derivative of #y=cos(x)/csc(x)#?

Question

Aurora Alvarado · Accepted Answer

There are two ways. The simpler way would be to recall that #csc(x) = 1/sin(x),# so #cos(x)/csc(x) = cos(x)sin(x)# [for all #x!=(n)pi# (for radians) or #x!=(n)180# (for degrees) where #n# is any integer]. We then use the product rule for differentiation, which says that for #f(x) = g(x)h(x), f'(x) = g'(x)h(x) + g(x)h'(x)# In the particular example provided, the restriction on the domain is required because, if #x=n*pi# (for radians) or #x=n*180# (for degrees), then #sin(x)=0 -> csc(x) = 1/sin(x) = 1/0#. However, the argument can be made that since #1/0=oo, cos(x)/(oo) = 0 = cos(x)sin(x) = cos(x)0# at these points, thus making the domain restriction unnecessary.  Given that we have changed our function into #y(x) = cos(x)sin(x)#, the next step is to find #y'(x)#. Assuming we know the derivatives of the sine and cosine functions, this is relatively straightforward.  #y'(x) = cos'(x)sin(x) + cos(x)sin'(x)#
#y'(x) = -sin(x)sin(x) + cos(x)cos(x)#
#y'(x) = -sin^2(x) + cos^2(x)# At this point, we could stop, but by using the trigonometric double-angle identities, we find further that... #y'(x) = -sin^2(x) + cos^2(x) = cos^2(x) - sin^2(x) = cos(2x)# Thus, for full simplification, we arrive at... #y'(x) = cos(2x)#.

Ezekiel Alexander · Answer

undefined To find the derivative of $ y = \frac{\cos(x)}{\csc(x)} $, you can use the quotient rule, which states that if you have a function in the form $ \frac{u(x)}{v(x)} $, then its derivative is given by $ \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $. Applying this rule to the given function, the derivative is:

$$ y' = \frac{(-\sin(x) \cdot \csc(x) - \cos(x) \cdot (-\csc(x)\cot(x)))}{(\csc(x))^2} $$

After simplifying, you can express the derivative as:

$$ y' = -\frac{\sin(x)\csc(x) + \cos(x)\cot(x)}{\sin^2(x)} $$