Understanding the Derivative of Cot(x): Step-by-Step Guide and Trigonometric Manipulations

Dx {cot x}=?

To find the derivative of cot(x), we’ll use the quotient rule:

The quotient rule states that for two functions u(x) and v(x), the derivative of their quotient u(x)/v(x) is given by:

[d(u(x)/v(x))] / dx = (v(x) * du(x)/dx – u(x) * dv(x)/dx)/(v(x))^2

In this case, u(x) = 1 and v(x) = tan(x).

Let’s calculate the derivatives of u(x) and v(x):

du(x)/dx = d(1)/dx = 0 (since the derivative of a constant is zero)

dv(x)/dx = d(tan(x))/dx

To find the derivative of tan(x), we can use the chain rule:

d(tan(x))/dx = sec^2(x) * dx/dx

dx/dx = 1 (since the derivative of x with respect to x is 1)

Therefore, dv(x)/dx = sec^2(x).

Now we can substitute these derivatives into the quotient rule:

[d(cot(x))] / dx = (tan(x) * 0 – 1 * sec^2(x))/(tan(x))^2

Simplifying, we have:

[d(cot(x))] / dx = -sec^2(x)/(tan(x))^2

Using the trigonometric identities, we know that:

(sec(x))^2 = 1 + (tan(x))^2

Re-arranging, we have:

(sec(x))^2 – (tan(x))^2 = 1

Dividing throughout by (tan(x))^2, we get:

(sec(x))^2/(tan(x))^2 – 1 = 1/(tan(x))^2

Simplifying the left side, we have:

1/(sin(x))^2 – 1 = 1/(tan(x))^2

Recall that the reciprocal of sine is cosecant, or csc(x), so we can substitute:

(csc(x))^2 – 1 = 1/(tan(x))^2

Substituting this expression into the derivative we found earlier:

[d(cot(x))] / dx = -[(csc(x))^2 – 1]

[d(cot(x))] / dx = -csc^2(x) + 1

So, the derivative of cot(x) is -csc^2(x) + 1.

Note: The above answer assumes that x is not equal to the zeroes of cot(x), where cot(x) is undefined.