Differentiate with respect to x
1/sin x
let y = 1/sinx
= (sinx)^-1
dy/dx = -1(sinx)^-2 (cosx)
= cosx (1/sin^2 x)
= cosx/sinx * 1/sinx
= cotx cscx
Thank you very much Reiny. I was so confused but i have understand from your solution
To differentiate 1/sin(x) with respect to x, we can use the quotient rule. The quotient rule states that for functions f(x) and g(x),
d/dx (f(x)/g(x)) = (g(x)*f'(x) - f(x)*g'(x))/(g(x))^2,
where f'(x) and g'(x) are the derivatives of f(x) and g(x) with respect to x, respectively.
In this case, f(x) = 1 and g(x) = sin(x). To differentiate 1/sin(x), we need to find the derivatives of f(x) and g(x) first.
f'(x) = d/dx (1) = 0 (since the derivative of a constant is always zero),
g'(x) = d/dx (sin(x)) = cos(x) (using the derivative formula for sine).
Now we can apply the quotient rule by substituting the values into the formula:
d/dx (1/sin(x)) = (sin(x)*0 - 1*cos(x))/(sin(x))^2.
Simplifying this expression, we get:
d/dx (1/sin(x)) = -cos(x)/(sin(x))^2.
Therefore, the derivative of 1/sin(x) with respect to x is -cos(x)/(sin(x))^2.