d/dx(cotx)
To find the derivative of cot(x) with respect to x, we can use the quotient rule
To find the derivative of cot(x) with respect to x, we can use the quotient rule. The cotangent function can be written as the cosine function divided by the sine function: cot(x) = cos(x)/sin(x).
Now, let’s use the quotient rule, which states that if we have a function in the form f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, the derivative of f(x) with respect to x is given by:
f'(x) = (g'(x) * h(x) – g(x) * h'(x)) / (h(x))^2
In our case, g(x) = cos(x) and h(x) = sin(x). We can find the derivatives of g(x) and h(x) as follows:
g'(x) = -sin(x)
h'(x) = cos(x)
Now, substituting these values into the quotient rule, we get:
cot'(x) = ( -sin(x) * sin(x) – cos(x) * cos(x) ) / (sin(x))^2
Simplifying the numerator gives:
cot'(x) = -sin^2(x) – cos^2(x) / sin^2(x)
Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can further simplify the numerator:
cot'(x) = -(1) / sin^2(x)
Finally, we can rewrite sin^2(x) as (1-cos^2(x)) using the Pythagorean identity sin^2(x) + cos^2(x) = 1:
cot'(x) = -(1) / (1 – cos^2(x))
Using the Pythagorean identity once more, cos^2(x) = 1 – sin^2(x), we can simplify further:
cot'(x) = -(1) / (1 – (1 – sin^2(x)))
cot'(x) = -(1) / (1 – 1 + sin^2(x))
cot'(x) = -(1) / sin^2(x)
Therefore, the derivative of cot(x) with respect to x is -1/sin^2(x), or equivalently, -csc^2(x).
More Answers:
[next_post_link]