In Calculus, differentiating inverse trigonometric functions is a core requirement for solving advanced integration and differential equations.
The mathematical formula for the derivative of the inverse cosine function, written as d/dx [cos⁻¹(x)] or d/dx [arccos(x)], is:
-1 / √(1 - x²) (where -1 < x < 1)
Function: y = cos⁻¹(x) [also written as arccos(x)].
Derivative Formula: d/dx [cos⁻¹(x)] = -1 / √(1 - x²).
Domain Restriction: The formula is only valid for -1 < x < 1. It is undefined at exactly 1 and -1 because the denominator becomes zero.
Memory Trick: It is identical to the derivative of sin⁻¹(x), but negative.
You don't just have to memorize it; you can easily prove it using implicit differentiation.
Step 1: Let y = cos⁻¹(x). Step 2: Therefore, x = cos(y). (The domain of y is restricted to 0 ≤ y ≤ π). Step 3: Differentiate both sides with respect to x using the chain rule: d/dx (x) = d/dx [cos(y)] 1 = -sin(y) * (dy/dx)
Step 4: Rearrange the equation to solve for dy/dx: dy/dx = -1 / sin(y)
Step 5: We know from the fundamental trigonometric identity that sin²(y) + cos²(y) = 1. Therefore, sin(y) = √(1 - cos²(y)). Since we established in Step 2 that cos(y) = x, we can substitute x into the equation: sin(y) = √(1 - x²).
Step 6: Substitute this back into the derivative equation from Step 4: dy/dx = -1 / √(1 - x²). (The proof is complete).
A very easy way to remember this formula is to compare it to the derivative of sin⁻¹(x). The derivative of sin⁻¹(x) is positive [1 / √(1 - x²)]. The derivative of cos⁻¹(x) is the exact same formula, just with a negative sign.
The derivative of cos inverse x with respect to x is -1 divided by the square root of (1 - x²).
They are exactly the same formula, except the derivative of cos inverse has a negative sign, while the derivative of sin inverse is positive.
Because if x = 1 or x = -1, the expression (1 - x²) becomes 0. The square root of zero is zero, and dividing by zero makes the derivative mathematically undefined.
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