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

2 Answers
Jul 5, 2017

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

Explanation:

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)#

Jul 5, 2017

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

Explanation:

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))#