Prove that d/dx(cot x)= -csc^2x
ctn= cos/sin
d/dt ctn= d/dx (cos/sin)=1/sin (d cos)+ cos(d sin^-1)
To prove that the derivative of the cotangent function is equal to minus the cosecant squared function, we can use the definition of the cotangent function and the derivative rules.
The cotangent function (cot x) is defined as the reciprocal of the tangent function (tan x), which can be expressed as cos x / sin x.
To find the derivative of cot x with respect to x (denoted as d/dx(cot x)), we can use the quotient rule. The quotient rule states that if we have a function f(x) = g(x) / h(x), then the derivative of f(x) with respect to x is given by:
d/dx(f(x)) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
Let's apply this rule to cot x:
g(x) = cos x and h(x) = sin x
g'(x) = -sin x (derivative of cos x)
h'(x) = cos x (derivative of sin x)
Substituting these values into the quotient rule formula:
d/dx(cot x) = (sin x * (-sin x) - cos x * cos x) / (sin x)^2
Simplifying further:
= (-sin^2x - cos^2x) / (sin^2x)
= -(sin^2x + cos^2x) / (sin^2x)
= -1 / (sin^2x)
= -csc^2x
Thus, we have proven that d/dx(cot x) = -csc^2x.