Understanding the Derivative of Cot(x) | Finding the Tangent to Optimize Your Math Skills

d/dx(cotx)

To find the derivative of cot(x) with respect to x, we will use the quotient rule

To find the derivative of cot(x) with respect to x, we will use the quotient rule. The quotient rule states that for a function u(x) = f(x)/g(x), where f(x) and g(x) are differentiable functions, the derivative is given by:

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

First, let’s express cot(x) as a quotient of sine and cosine:

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

Now, applying the quotient rule, let f(x) = cos(x) and g(x) = sin(x):

f'(x) = -sin(x) (using the derivative of cosine)
g'(x) = cos(x) (using the derivative of sine)

Now substituting into the quotient rule formula:

[d/dx(cot(x))] = [(cos(x) * sin(x)) – (cos(x) * (-sin(x)))] / [sin(x)]^2
= [cos(x) * (sin(x) + sin(x))] / [sin(x)]^2
= 2cos(x) / sin(x)

Since cot(x) can also be expressed as 1/tan(x), we can simplify further:

[d/dx(cot(x))] = 2cos(x) / sin(x)
= 2cos(x) / cos(x) / sin(x) / cos(x)
= 2 / tan(x)

Therefore, the derivative of cot(x) with respect to x is 2 / tan(x).

More Answers:

[next_post_link]