How do you differentiate #f(x) = (sinx)/(1-cosx)# using the quotient rule?

Question

Paisley Johnson · Accepted Answer

#f'(x) = 1/(cos(x) - 1)#

For derivatives, the Quotient Rule stipulates:

If #f(x) = g(x)/(h(x))#

then

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

See also the Quotient Rule article at https://tutor.hix.ai

So, the first step is to identify #g(x)# and #h(x)# in your original function:

#f(x) = sin(x)/(1- cos(x))#

so

#g(x) = sin(x)#

#h(x) = 1- cos(x)#

We then differentiate the two components of #f(x)# with respect to #x#:

#g'(x) = cos(x)#

#h'(x) = -(-sin(x)) = sin(x)#

Using the Quotient Rule as a guide, let's put everything together and then simplify:

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

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

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

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

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

#f'(x) = (cos(x)- 1)/((-1)(1- cos(x))(-1)(1- cos(x)))#

#f'(x) = cancel(cos(x) - 1)/(cancel((cos(x) - 1))*(cos(x) - 1)#

#f'(x) = 1/(cos(x) - 1)#

Cameron Alexander · Answer

undefined To differentiate $ f(x) = \frac{\sin x}{1 - \cos x} $ using the quotient rule, follow these steps:

1. Apply the quotient rule, which states that for functions $ u(x) $ and $ v(x) $, the derivative of $ \frac{u(x)}{v(x)} $ is given by $ \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $.

2. Identify $ u(x) = \sin x $ and $ v(x) = 1 - \cos x $.

3. Compute the derivatives $ u'(x) $ and $ v'(x) $, which are $ u'(x) = \cos x $ and $ v'(x) = \sin x $.

4. Apply the quotient rule formula: 
$$ f'(x) = \frac{(u'(x)v(x) - u(x)v'(x))}{(v(x))^2} $$
$$ = \frac{(\cos x)(1 - \cos x) - (\sin x)(-\sin x)}{(1 - \cos x)^2} $$

5. Simplify the expression to obtain the derivative $ f'(x) $.

$$ f'(x) = \frac{\cos x - \cos^2 x + \sin^2 x}{(1 - \cos x)^2} $$

$$ = \frac{\cos x - \cos^2 x + (1 - \cos^2 x)}{(1 - \cos x)^2} $$

$$ = \frac{\cos x + 1 - 2\cos^2 x}{(1 - \cos x)^2} $$

$$ = \frac{\cos x + 1 - 2(1 - \sin^2 x)}{(1 - \cos x)^2} $$

$$ = \frac{\cos x + 1 - 2 + 2\sin^2 x}{(1 - \cos x)^2} $$

$$ = \frac{\sin^2 x + \cos x - 1}{(1 - \cos x)^2} $$

Ezekiel Alexander · Answer

undefined To differentiate $ f(x) = \frac{\sin(x)}{1 - \cos(x)} $ using the quotient rule:

1. Apply the quotient rule formula, $ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} $.
2. Identify $ u = \sin(x) $ and $ v = 1 - \cos(x) $.
3. Compute the derivatives $ u' $ and $ v' $:
   - $ u' = \cos(x) $ (derivative of sine function)
   - $ v' = \sin(x) $ (derivative of $ 1 - \cos(x) $)
4. Substitute $ u, v, u', $ and $ v' $ into the quotient rule formula.
5. Simplify the expression.