I don't know to find the derivative of the following trig functions. Please help.
f(x)= cotx
h(x)= secx
i(x)= cscx
any calculus text will list the trig functions and their derivatives.
To find the derivatives of trigonometric functions, we can use the rules of differentiation and the derivatives of basic trigonometric functions. Here's how to find the derivatives of the given trigonometric functions:
1. Derivative of f(x) = cot(x):
The cotangent function can be expressed as cot(x) = cos(x) / sin(x). To find the derivative, we can use the quotient rule. The quotient rule states that d/dx(u(x)/v(x)) = [v(x)*u'(x) - u(x)*v'(x)] / [v(x)]^2.
In this case, u(x) = cos(x) and v(x) = sin(x). Let's differentiate them:
u'(x) = -sin(x) (derivative of cos(x))
v'(x) = cos(x) (derivative of sin(x))
Now, we can plug these values into the quotient rule formula:
f'(x) = [sin(x) * (-sin(x)) - cos(x) * cos(x)] / sin^2(x)
= [-sin^2(x) - cos^2(x)] / sin^2(x)
= -1 / sin^2(x)
= -csc^2(x)
Therefore, the derivative of f(x) = cot(x) is f'(x) = -csc^2(x).
2. Derivative of h(x) = sec(x):
The secant function can be written as sec(x) = 1 / cos(x). To find the derivative, we can use the quotient rule again. Let's label u(x) = 1 and v(x) = cos(x):
u'(x) = 0 (the derivative of a constant is 0)
v'(x) = -sin(x) (derivative of cos(x))
Applying the quotient rule, we have:
h'(x) = [cos(x) * 0 - 1 * (-sin(x))] / cos^2(x)
= sin(x) / cos^2(x)
= sin(x) * sec^2(x)
Therefore, the derivative of h(x) = sec(x) is h'(x) = sin(x) * sec^2(x).
3. Derivative of i(x) = csc(x):
The cosecant function can be expressed as csc(x) = 1 / sin(x). Again, we'll apply the quotient rule. Taking u(x) = 1 and v(x) = sin(x):
u'(x) = 0 (the derivative of a constant is 0)
v'(x) = cos(x) (derivative of sin(x))
Using the quotient rule:
i'(x) = [sin(x) * 0 - 1 * cos(x)] / sin^2(x)
= -cos(x) / sin^2(x)
= -cos(x) * csc^2(x)
Thus, the derivative of i(x) = csc(x) is i'(x) = -cos(x) * csc^2(x).
Hope this helps!