Why is the derivative of cot x equal to -cosec^2 x instead of -1/tan^2?

[tomwilliam]

Can someone show me why the derivative of

k(x) = cot x

Is

k'(x) = -cosec^2 x

And not

k'(x) = -1/tan^2

Thanks!

 

[Char. Limit]

Well, cot(x) is really the quotient of two functions: cos(x)/sin(x). Then, using the quotient rule...

$$\frac{d cot(x)}{dx} = \frac{(sin(x))(-sin(x)) - (cos(x))(cos(x))}{sin^2(x)}$$

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

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

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

 

[JCVD]

One way to derive the formula for the derivative of cot(x) is to rewrite it as 1/tan(x) (or cos(x)/sin(x)) and use the quotient rule. The mistake you probably made in getting -1/tan^2(x) was to treat cot(x) as the composition of the functions y=1/u(x) and u(x)=tan(x) and then forgetting to apply the chain rule: you must multiply -1/tan^2(x) by the derivative of tan(x).

 

[tomwilliam]

Thanks to you both. That's very clear.

 

[tomwilliam]

Just for information, the mistaken thought process I had was to differentiate tan^-1 as -1* tan^-2...

 

[Mute]

Just for information, the mistaken thought process I had was to differentiate tan^-1 as -1* tan^-2...

You can do that, but your calculation is incomplete. You have to use the chain rule:

$$\frac{d}{dx}(f(g(x))) = \frac{df(g)}{dg} \frac{dg(x)}{dx}$$

With f(g) = 1/g and g = tan(x), you should again find that the derivative of cot(x) is -csc(x).

 

[Mark44]

Just for information, the mistaken thought process I had was to differentiate tan^-1 as -1* tan^-2...

As Mute pointed out, this will work if you use the chain rule correctly. However, the expression tan^-1(x) usually is taken to mean the inverse of the tangent function (AKA the arctangent), not its reciprocal.

If you write d/dx(cot(x)) = d/dx(tan^-1(x)), you are really going to confuse a lot of people.

 

[tomwilliam]

Thankyou.