derivative of sin⁻¹x
derivative of sin⁻¹x is equal to 1 divided by the square root of (1-x²).
The derivative of sin⁻¹x can be determined by using the formula for the derivative of inverse functions, which is:
d/dx [f⁻¹(x)] = 1 / f ‘(f ⁻¹(x))
Applying this formula to sin⁻¹x, we get:
d/dx [sin⁻¹x] = 1 / cos (sin⁻¹x)
To simplify this expression, we can use the Pythagorean identity for the sine and cosine functions, which is:
sin²θ + cos²θ = 1
Rearranging this equation, we get:
cos²θ = 1 – sin²θ
Substituting θ = sin⁻¹x into this equation, we get:
cos²(sin⁻¹x) = 1 – sin²(sin⁻¹x)
Using the definition of the inverse sine function, we can rewrite sin(sin⁻¹x) as x, and cos(sin⁻¹x) as √(1-x²). Therefore,
cos²(sin⁻¹x) = 1 – x²
Substituting this expression into the previous equation for the derivative of sin⁻¹x, we get:
d/dx [sin⁻¹x] = 1 / √(1-x²)
Therefore, the derivative of sin⁻¹x is equal to 1 divided by the square root of (1-x²).
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