In Class 12 Calculus, finding the derivative of inverse trigonometric functions is a standard requirement. The function sin⁻¹(x) (also written as arcsin(x)) has a very specific and frequently used derivative formula.
The derivative of cos⁻¹(x) is exactly the same as sin⁻¹(x), except it has a negative sign: -1 / √(1 - x²).
The derivative of sin⁻¹(x) with respect to x is:
d/dx [sin⁻¹(x)] = 1 / √(1 - x²)
(Note: This formula is valid only for -1 < x < 1).
We can prove this using implicit differentiation.
Step 1: Let y = sin⁻¹(x) Step 2: This means, sin(y) = x Step 3: Differentiate both sides with respect to x using the chain rule: d/dx [sin(y)] = d/dx [x] cos(y) · (dy/dx) = 1 Step 4: Rearrange to solve for dy/dx: dy/dx = 1 / cos(y) Step 5: We know from the identity sin²y + cos²y = 1, that cos(y) = √(1 - sin²y). Since sin(y) = x, then cos(y) = √(1 - x²). Step 6: Substitute this back: dy/dx = 1 / √(1 - x²).
The domain is restricted because if x = 1 or x = -1, the denominator √(1 - x²) becomes zero, making the derivative undefined (tending to infinity).
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