In Calculus (Class 12 Math), finding the derivatives of inverse trigonometric functions is a standard requirement. The function sin⁻¹(x) is also written as arcsin(x).
Formula: d/dx(sin⁻¹x) = 1 / √(1 - x²).
Also written as: arcsin(x).
Domain: Valid for x between -1 and 1 (exclusive).
Proof method: Implicit differentiation using chain rule.
The derivative of sin⁻¹(x) with respect to x is:
d/dx [sin⁻¹(x)] = 1 / √(1 - x²)
Condition: This is valid for -1 < x < 1 (because the denominator cannot be zero, and the domain of sin⁻¹ is [-1, 1]).
We can prove this using implicit differentiation.
Step 1: Let y = sin⁻¹(x) This means: sin(y) = x (where -π/2 ≤ y ≤ π/2)
Step 2: Differentiate both sides with respect to x: d/dx [sin(y)] = d/dx [x] By chain rule: cos(y) · (dy/dx) = 1
Step 3: Solve for dy/dx: dy/dx = 1 / cos(y)
Step 4: Convert cos(y) back into terms of x. We know from the Pythagorean identity: sin²(y) + cos²(y) = 1 Therefore: cos(y) = √(1 - sin²(y)) Since we defined x = sin(y) in Step 1, substitute x: cos(y) = √(1 - x²)
Step 5: Final Substitution: dy/dx = 1 / √(1 - x²)
Hence proved.
The derivative of sin inverse x (sin⁻¹x) is 1 divided by the square root of (1 - x²).
You prove it by setting y = arcsin x, rewriting it as sin y = x, differentiating both sides using the chain rule to get cos y (dy/dx) = 1, and then using the trigonometric identity to replace cos y with √(1 - x²).
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