d/dx[sin^-1(x)]=
1/sqrt(1-x^2)
To find the derivative of sin^-1(x), we first need to use the inverse property of trigonometric functions and express sin^-1(x) in terms of the inverse sine function:
sin(sin^-1(x)) = x
Now, we take the derivative of both sides of this equation with respect to x:
cos(sin^-1(x)) * d/dx[sin^-1(x)] = 1
Here, we have used the chain rule on the left-hand side of the equation. Now, we can solve for d/dx[sin^-1(x)]:
d/dx[sin^-1(x)] = 1 / cos(sin^-1(x))
To simplify this expression further, we use the Pythagorean identity: sin^2(x) + cos^2(x) = 1. Since sin^-1(x) represents an angle whose sine is x, we can write:
sin^-1(x) + cos^2(sin^-1(x)) = 1
Solving for cos(sin^-1(x)), we get:
cos(sin^-1(x)) = sqrt(1 – sin^2(sin^-1(x))) = sqrt(1 – x^2)
Substituting this into our expression for the derivative, we get:
d/dx[sin^-1(x)] = 1 / sqrt(1 – x^2)
Therefore, the derivative of sin^-1(x) is 1 / sqrt(1 – x^2).
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