How Does the Quotient Rule Apply to Derivatives of Cotangent Functions?

[Torshi]

Homework Statement

Use quotient rule

Homework Equations

d/dx (cot(x)

The Attempt at a Solution

d/dx (cot(x))
= cos(x)/sin(x)
= (sin(x))(-sin(x)) - (cos(x))(cos(x)) / (sin(x))^2
= -(sin(x))^2 - (cos(x))^2 / (sin(x))^2
= -1 - (cos(x))^2


Does the last step equal 1 + (cos(x))^2 and how come that turns into negative -(csc(x))^2

 

[jbunniii]

Homework Statement

Use quotient rule

Homework Equations

d/dx (cot(x)

The Attempt at a Solution

d/dx (cot(x))
= cos(x)/sin(x)
= (sin(x))(-sin(x)) - (cos(x))(cos(x)) / (sin(x))^2
= -(sin(x))^2 - (cos(x))^2 / (sin(x))^2
= -1 - (cos(x))^2


Does the last step equal 1 + (cos(x))^2 and how come that turns into negative -(csc(x))^2

I think you dropped some parentheses and this led you to a wrong answer. That first step should be:
$$ \frac{d}{dx} \cot(x) = \frac{\sin(x)(-\sin(x)) - \cos(x)\cos(x)}{\sin^2(x)}$$
Now simplify the numerator and see what you end up with.

 

[Torshi]

I think you dropped some parentheses and this led you to a wrong answer. That first step should be:
$$ \frac{d}{dx} \cot(x) = \frac{\sin(x)(-\sin(x)) - \cos(x)\cos(x)}{\sin^2(x)}$$
Now simplify the numerator and see what you end up with.

= (-1)-(cos(x))^2

 

[jbunniii]

= (-1)-(cos(x))^2

Shouldn't the second term be divided by $\sin^2(x)$?

 

[Torshi]

Shouldn't the second term be divided by $\sin^2(x)$?

Hold on
-1 - (cos(x))^2 / (sin(x))^2 = -1/(sin(x))^2 = -(csc(x))^2

but...

-(sin(x))^2 - (cos(x))^2 / (sin(x))^2 = 1/(sin(x))^2 = -(csc(x))^2

 

[jbunniii]

Hold on
-1 - (cos(x))^2 / (sin(x))^2 = -1/(sin(x))^2 = -(csc(x))^2

How did you get that?

but...

-(sin(x))^2 - (cos(x))^2 / (sin(x))^2 = 1/(sin(x))^2 = -(csc(x))^2

I think there are some parentheses missing again. The expression on the left should be
$$\frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)}$$
Now you can use the identity $\sin^2(x) + \cos^2(x) = 1$ in the numerator.

 

[Torshi]

How did you get that?

I think there are some parentheses missing again. The expression on the left should be
$$\frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)}$$
Now you can use the identity $\sin^2(x) + \cos^2(x) = 1$ in the numerator.

My main question is do both of the following equations equal -(csc(x))^2

1/(sin(x))^2 and -1/(sin(x))^2

or which one of those two define -(csc(x))^2 ?

 

[HallsofIvy]

Since, by definition, csc(x)= 1/sin(x), it follows that sin(x)= 1/csc(x) and then that sin^2(x)= 1/csc^2(x).

If you want "-" on the right, you will have have a "-" on the right.

 

[Mark44]

Homework Statement

Use quotient rule

Homework Equations

d/dx (cot(x)

The Attempt at a Solution

d/dx (cot(x))
= cos(x)/sin(x)

You haven't actually taken the derivative yet, so the above should be:
=d/dx([/color]cos(x)/sin(x))[/color]

Others have already commented about the missing parentheses and the incorrect parts.

= (sin(x))(-sin(x)) - (cos(x))(cos(x)) / (sin(x))^2
= -(sin(x))^2 - (cos(x))^2 / (sin(x))^2
= -1 - (cos(x))^2


Does the last step equal 1 + (cos(x))^2 and how come that turns into negative -(csc(x))^2