Derivative of cscx
To find the derivative of csc(x), we can start by expressing it in terms of sine and cosine
To find the derivative of csc(x), we can start by expressing it in terms of sine and cosine. Recall that csc(x) is the reciprocal of sin(x), so we have:
csc(x) = 1/sin(x)
Now, let’s differentiate both sides of the equation with respect to x:
d/dx (csc(x)) = d/dx (1/sin(x))
To simplify the differentiation, we can rewrite the right side using the quotient rule. The quotient rule states that for functions u(x) and v(x), the derivative of u(x)/v(x) is given by:
d/dx (u(x)/v(x)) = (v(x) * du(x)/dx – u(x) * dv(x)/dx) / (v(x))^2
Applying the quotient rule to our equation, we have:
d/dx (csc(x)) = [(sin(x) * d/dx(1)) – (1 * d/dx(sin(x)))] / (sin(x))^2
Now, let’s find the derivatives of the individual terms.
d/dx(1) = 0 (since the derivative of a constant is zero)
Using the chain rule, we have:
d/dx(sin(x)) = cos(x)
Plugging these values back into the equation, we get:
d/dx (csc(x)) = [(sin(x) * 0) – (1 * cos(x))] / (sin(x))^2
Simplifying further, we have:
d/dx (csc(x)) = -cos(x) / (sin(x))^2
Therefore, the derivative of csc(x) is -cos(x) / (sin(x))^2.
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