Differentiate sin^-1x

Inverse trigonometric functions are the inverse ratios of the basic trigonometric ratios.

Answer: The differentiation of sin^-1x is 1/⎷(1-x²).

Let us see how to solve it.

Explanation:

Let's assume f(x) = sin^-1x.

Substitute x = sin Ø in equation f(x) =  sin^-1x,

f(sin Ø)=sin^-1sin Ø = Ø

Now differentiate the equation f(sin Ø) = Ø with respect to Ø,

f(sin Ø) = Ø

f^'(sin Ø)cos Ø = 1

f^'(sin Ø) = 1/cos Ø

As sin Ø = x, so to calculate the value of cos Ø use the formula, cos²Ø = 1 - sin²Ø.

cos²Ø = 1 - sin²Ø

cos²Ø = 1 - x²

cos Ø = ⎷(1 - x²)

Now, substitute cos Ø = ⎷(1 - x²) and sin Ø = x in the equation f^'(sin Ø) = 1/cos Ø and simplify.

f^'(sin Ø) = 1/cos Ø

f^'(x) =  1/⎷(1 - x²) 

Thus, the differentiation of  sin^-1x is  1/⎷(1 - x²).