How do you differentiate #f(x) = (cosx)/(sinx)# using the quotient rule?

Question

Samantha Adkison · Accepted Answer

#f'(x) = -csc^2x#

Apply the rule of quotient.

Let #f(x) = (g(x))/(h(x))#, then #g(x) = cosx# and #h(x) = sinx#. Then #g'(x) = -sinx# and #h'(x) = cosx#.

#f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2#

#f'(x) = (-sinx(sinx) - cosx(cosx))/(sinx)^2#

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

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

#f'(x) = -1/sin^2x#

#f'(x) = -csc^2x#

I hope this is useful!

Isabella Ackerman · Answer

undefined To differentiate $ f(x) = \frac{\cos(x)}{\sin(x)} $ using the quotient rule:

1. Apply the quotient rule, which states: 
$$ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

2. Identify $ u(x) = \cos(x) $ and $ v(x) = \sin(x) $.

3. Compute the derivatives:
   - $ u'(x) = -\sin(x) $
   - $ v'(x) = \cos(x) $

4. Apply the quotient rule formula:
$$ f'(x) = \frac{(-\sin(x))(\sin(x)) - (\cos(x))(\cos(x))}{(\sin(x))^2} $$

5. Simplify the expression:
$$ f'(x) = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)} $$

6. Use the trigonometric identity $ \sin^2(x) + \cos^2(x) = 1 $:
$$ f'(x) = \frac{-1}{\sin^2(x)} $$

7. Rewrite the expression in terms of $\csc(x)$:
$$ f'(x) = -\csc^2(x) $$