How do you differentiate #f(x)=cotx/(1-sinx)#?

Question

Scarlett Allison · Accepted Answer

#d/(dx) ((cotx)/(1-sinx)) = color(blue)(((cotx)(cosx) + (-csc^2x)(1-sinx))/((1-sinx)^2)#

We can use the quotient rule:

#d/(dx) (u/v) = (v(du)/(dx) - u(dv)/(dx))/(v^2)#

where

#u = cotx#

and

#v = 1-sinx#

#(-cotx(d/(dx)(1-sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)#

#(-cotx(d/(dx)(1)-d/(dx)(sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)#

The derivative of #1# is #0#:

#(-cotx(-d/(dx)(sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)#

#(cotx(d/(dx)(sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)#

The derivative of #sinx# is #cosx#:

#((cotx)(cosx) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)#

The derivative of #cotx# is #-csc^2x#:

#color(blue)(((cotx)(cosx) + (-csc^2x)(1-sinx))/((1-sinx)^2)#

or

#color(blue)(((cotx)(cosx))/((1-sinx)^2) - (csc^2x)/(1-sinx)#

Charles Anctil · Answer

#f'(x) = (cos(x)cot(x))/(1-sin(x))^2-csc^2(x)/(1-sin(x))# In order to derive #cot(x)/(1-sin(x))#, you need to apply the quotient rule first. The quotient rule states this: #(f'(x)g(x) - g'(x)f(x))/g(x)^2# Let's assume #(1-sin(x))# is #g(x)#. Now, the #f(x)# that we will use for the quotient rule is not the equation they give us. It is the numerator, which is #cot(x)#. So let's plug in our newly found #f(x)# and #g(x)#, which is: #(d/dxcot(x)(1-sin(x))-d/dx(1-sin(x))(cot(x)))/(1-sin(x))^2# Now, the derivative of #cot(x)# is #-csc^2(x)# and the derivative of #(1-sin(x))# is #-cos(x)#. After finding the derivatives of #f(x)# and #g(x)#, we rewrite our previous equation: #(-csc^2(x)(1-sin(x))-(-cos(x))(cot(x)))/(1-sin(x))^2# Now we split the numerator into two parts: #(cot(x)cos(x))/(1-sin(x))^2-(csc^2(x)(1-sin(x)))/(1-sin(x))^2# For #(csc^2(x)(1-sin(x)))/(1-sin(x))^2# we can cancel out one #(1-sin(x))# from the numerator and denominator, so this simplifies to: #(csc^2(x))/(1-sin(x))# So the final answer is: #f'(x) = (cos(x)cot(x))/(1-sin(x))^2-csc^2(x)/(1-sin(x))#

Aurora Alvarado · Answer

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

$$ f'(x) = \frac{u'v - uv'}{v^2} $$

where $ u = \cot(x) $ and $ v = 1 - \sin(x) $. Differentiating $ u $ and $ v $:

$$ u' = -\csc^2(x) $$
$$ v' = -\cos(x) $$

Substituting into the quotient rule formula:

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

Simplify the numerator and denominator.