## d/dx (cot x)

### To find the derivative of the cotangent function, we can use the quotient rule

To find the derivative of the cotangent function, we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, then the derivative of f(x) is given by:

f'(x) = (g'(x) * h(x) – g(x) * h'(x))/[h(x)]^2

In this case, we have f(x) = cot(x), and we know that cot(x) is defined as the ratio of cosine(x) to sine(x), i.e., cot(x) = cos(x)/sin(x). Therefore, we can rewrite cot(x) as f(x) = cos(x)/sin(x).

Applying the quotient rule to f(x), we have:

f'(x) = [(cos(x) * sin(x))’ * sin(x) – cos(x) * (sin(x))’]/[sin(x)]^2

Let’s find the derivatives of the individual components:

Derivative of cos(x): Recall that the derivative of cosine is given by -sin(x). Therefore, (cos(x))’ = -sin(x).

Derivative of sin(x): The derivative of sine is found to be cosine. So, (sin(x))’ = cos(x).

Plugging these derivatives back into the quotient rule, we have:

f'(x) = [(-sin(x) * sin(x)) * sin(x) – cos(x) * cos(x)]/[sin(x)]^2

Simplifying further:

f'(x) = [-sin^2(x)*sin(x) – cos^2(x)]/[sin^2(x)]

Now, we can apply some trigonometric identities to simplify the expression:

Recall that 1 – cos^2(x) = sin^2(x). Rearranging this equation, we get: cos^2(x) = 1 – sin^2(x).

Substituting this identity into our expression, we have:

f'(x) = [-sin^2(x)*sin(x) – (1 – sin^2(x))]/[sin^2(x)]

Expanding and simplifying:

f'(x) = [-sin^3(x) + sin^2(x) – 1]/[sin^2(x)]

This is the derivative of cot(x), given by f'(x) = [-sin^3(x) + sin^2(x) – 1]/[sin^2(x)].

Note: It is important to remember that the domain of cot(x) is all real numbers except for the vertical asymptotes, which occur when sin(x) = 0.

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