derivative of 1/(sin(x))
To simplify differentiation, we rewrite 1sin(x) as sin(x)-1
Let's differentiate: ddx(sin(x)-1)
Apply the differentiation rule: ddx(fn) = n·ddx(f)·fn -1 with f = sin(x)
ⓘ The derivative of x^n is n*x^(n-1): ddx(sin(x)-1) = -1·ddx(sin(x))·sin(x)-2
-1·ddx(sin(x))·sin(x)-2
ⓘ Multiplying by -1 gives the opposite: -1·ddx(sin(x))·sin(x)-2 = -ddx(sin(x))·sin(x)-2
-ddx(sin(x))·sin(x)-2
ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base: -ddx(sin(x))·sin(x)-2 = -ddx(sin(x))sin(x)2
-ddx(sin(x))sin(x)2
ⓘ The derivative of sin(x) is cos(x): ddx(sin(x)) = cos(x)
▶Derivative:   -cos(x)sin(x)2