# Differentiate sin^{-1}x

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

## Answer: The differentiation of sin^{-1}x is 1/⎷(1-x^{2}).

Let us see how to solve it.

**Explanation:**

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

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

f(sin Ø)=sin^{-1}sin 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, cos^{2}Ø = 1 - sin^{2}Ø.

cos^{2}Ø = 1 - sin^{2}Ø

cos^{2}Ø = 1 - x^{2}

cos Ø = ⎷(1 - x^{2})

Now, substitute cos Ø = ⎷(1 - x^{2}) 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 - x^{2})

### Thus, the differentiation of sin^{-1}x is 1/⎷(1 - x^{2}).

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