Exploring the derivative of csc(x) using the chain rule and trigonometric identities

Derivative of csc x

To find the derivative of csc(x), we can use the chain rule

To find the derivative of csc(x), we can use the chain rule. Recall that the derivative of the reciprocal function, 1/x, is -1/x^2.

The chain rule states that if we have a composite function, y = f(g(x)), then the derivative dy/dx is given by dy/dx = f'(g(x)) * g'(x).

In this case, we have y = csc(x), which can be written as y = (1/sin(x)). Thus, we can treat csc(x) as a composition of two functions:
f(u) = 1/u, and
g(x) = sin(x).

To find the derivative dy/dx, let’s apply the chain rule:
dy/dx = f'(g(x)) * g'(x).

First, let’s find f'(u) by taking the derivative of f(u) = 1/u. Applying the derivative rule for the reciprocal function, we have:
f'(u) = -1/u^2.

Next, let’s find g'(x), the derivative of sin(x). The derivative of sin(x) is cos(x), so we have:
g'(x) = cos(x).

Now, we can substitute these results into the chain rule formula:
dy/dx = f'(g(x)) * g'(x)
= (-1/(sin(x))^2) * cos(x).

Simplifying the expression, we have:
dy/dx = -cos(x) / sin^2(x).

Alternatively, we can rewrite the final result using the fundamental trigonometric identity:
dy/dx = -1 / (sin(x) * sin(x) * cot(x)).

Therefore, the derivative of csc(x) is -cos(x) / sin^2(x) or -1 / (sin(x) * sin(x) * cot(x)).

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