I don't know to find the derivative of the following trig functions. Please help.

f(x)= cotx

h(x0= secx

i(x)= cscx

To find the derivatives of the given trigonometric functions, you can use the rules of differentiation along with the derivatives of basic trigonometric functions.

1. f(x) = cot(x):
To find the derivative of cot(x), you can use the quotient rule.
The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), then the derivative of f(x) can be found using the formula: f'(x) = (g'(x)h(x) - g(x)h'(x))/[h(x)]^2.

In the case of f(x) = cot(x), g(x) = 1 and h(x) = tan(x).
Therefore, applying the quotient rule, we have:
f'(x) = (0 * tan(x) - 1 * sec^2(x))/[tan(x)]^2
= -sec^2(x)/[tan(x)]^2
= -cos^2(x)/sin^2(x)
= -cos^2(x)/[1 - cos^2(x)]
= -cos^2(x)/(1 - cos^2(x))
= -1/(1 - cos^2(x))
= -1/sin^2(x)

Therefore, the derivative of f(x) = cot(x) is f'(x) = -1/sin^2(x).

2. h(x) = sec(x):
To find the derivative of sec(x), you can use the quotient rule as well, combined with the chain rule.
The chain rule states that if you have a function of the form f(g(x)), then the derivative of f(g(x)) can be found by multiplying the derivative of f(g) by the derivative of g(x).

In the case of h(x) = sec(x), g(x) = x and f(u) = sec(u).
Therefore, applying the chain rule, we have:
h'(x) = [d(sec(u))/du] * [du/dx]
= [sec(u)tan(u)] * [1]
= sec(x)tan(x)

Therefore, the derivative of h(x) = sec(x) is h'(x) = sec(x)tan(x).

3. i(x) = csc(x):
To find the derivative of csc(x), you can use the quotient rule and the chain rule, similar to the previous cases.
The chain rule states that if you have a function of the form f(g(x)), then the derivative of f(g(x)) can be found by multiplying the derivative of f(g) by the derivative of g(x).

In the case of i(x) = csc(x), g(x) = x and f(u) = csc(u).
Therefore, applying the chain rule, we have:
i'(x) = [d(csc(u))/du] * [du/dx]
= [-csc(u)cot(u)] * [1]
= -csc(x)cot(x)

Therefore, the derivative of i(x) = csc(x) is i'(x) = -csc(x)cot(x).

You can verify these results by comparing them with the trigonometric identity for derivatives, where the derivative of cot(x), sec(x), and csc(x) are equal to -csc^2(x), sec(x)tan(x), and -csc(x)cot(x), respectively.