Derivative of asin(x)

Answer

1−x2+1\frac{1}{\sqrt{-x^{2}+1}}
sqrt(-x^2+1)^(-1)

Derivative of asin(x)

Rule Used
Inverse Trigonometric Rule
Steps
  1. Recognize an inverse trigonometric function. It is not the reciprocal of the trig function: arcsin⁡(x)\arcsin(x) means arcsin⁡(x)\arcsin(x), the angle whose sine is x.
  2. The derivative comes from implicit differentiation of the inverse relation, then a Pythagorean identity that removes the angle.
  3. Final answer: −x2+1−1\sqrt{-x^2+1}^{-1}

d/dx arcsin(x) = d/dx sin^-1(x) = 1 / sqrt(1 - x^2), for -1 < x < 1.

Instant step-by-step solution for asin(x)

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Derivative of asin(x) = sqrt(-x^2+1)^(-1) Solve step-by-step for free: https://derivativecalculatorai.com/derivative-of-arcsin-x

Formula for arcsin(x) and sin^-1(x)

d/dx arcsin(x) = d/dx sin^-1(x) = 1 / sqrt(1 - x^2), for -1 < x < 1.

At x = ±1 the graph of arcsin has a vertical tangent, so the derivative is not a real number there. The calculator on this page is already filled with asin(x), which is the same function.

Open arcsin(x) in the derivative calculator

Implicit differentiation that produces 1/sqrt(1-x^2)

Let y = arcsin(x). By definition y is the unique angle in [-π/2, π/2] whose sine is x, so sin(y) = x. Differentiating that equation is shorter than differentiating the inverse from the limit definition.

The left side is a composition. The chain rule gives cos(y) · y'. The right side is the line y = x, whose derivative is 1. Canceling is then one division: y' = 1/cos(y), as long as cos(y) is not zero.

Cosine is non-negative on the range of arcsin. That fact picks the positive square root: cos(y) = sqrt(1 - sin^2(y)) = sqrt(1 - x^2). Substituting it in is the whole proof. The same line is the derivative of sin^-1(x); the notation does not change the function.

  1. Set y = arcsin(x), so sin(y) = x and y is in [-π/2, π/2].
  2. Differentiate both sides: cos(y) · dy/dx = 1.
  3. Solve for the derivative: dy/dx = 1/cos(y).
  4. Because y stays in [-π/2, π/2], cos(y) ≥ 0, so cos(y) = sqrt(1 - sin^2(y)).
  5. Replace sin(y) with x: cos(y) = sqrt(1 - x^2).
  6. Conclude d/dx arcsin(x) = 1/sqrt(1 - x^2), for -1 < x < 1.

Worked examples, including arcsin of a multiple

Slope of arcsin at the origin

  1. Substitute x = 0 into 1/sqrt(1 - x^2).
  2. The inside of the root is 1, and sqrt(1) = 1.
  3. The slope is 1. Near zero, arcsin(x) and the line y = x share a tangent.

Answer: 1

Solve this example

Derivative of arcsin(2x)

  1. The outer derivative is 1/sqrt(1 - u^2) with u = 2x.
  2. The inner derivative is 2. The chain rule multiplies them.
  3. 1 - (2x)^2 = 1 - 4x^2, so the result is 2/sqrt(1 - 4x^2).
  4. The domain shrinks: |2x| < 1, which means |x| < 1/2.

Answer: 2 / sqrt(1 - 4x^2), for |x| < 1/2

Solve this example

Derivative of x · arcsin(x)

  1. This is a product. Use (uv)' = u'v + uv' with u = x and v = arcsin(x).
  2. u' = 1 and v' = 1/sqrt(1 - x^2).
  3. The two terms are arcsin(x) and x/sqrt(1 - x^2). Both stay in the answer.

Answer: arcsin(x) + x/sqrt(1 - x^2)

Solve this example

Mistakes that turn arcsin into cosecant

Reading sin^-1(x) as 1/sin(x)

sin^-1(x) is arcsin(x). The reciprocal 1/sin(x) is csc(x), and its derivative is -csc(x) cot(x). That answer has vertical asymptotes at every multiple of π, which arcsin does not have.

Dropping the chain-rule factor on arcsin(2x)

1/sqrt(1 - 4x^2) is missing the derivative of the inside. The factor 2 has to be squared inside the root and also multiplied outside. Squaring only the x, as in 1 - 2x^2, is the other version of the same slip.

Taking the negative square root

cos(y) = ±sqrt(1 - x^2). On the range of arcsin the cosine is positive, so the derivative of arcsin is positive for every x in (-1, 1). The negative formula is the derivative of arccos(x).

Plugging in the endpoints

At x = 1 and x = -1 the expression 1/sqrt(1 - x^2) divides by zero. Arcsin is defined at those points and the derivative is not. A numerical check at x = 0.999 gets large; it does not settle to a finite slope.

Related functions

Frequently asked questions

What is the derivative of sin^-1(x)?

The derivative of sin^-1(x) is 1/sqrt(1 - x^2), for x strictly between -1 and 1. sin^-1(x) is another way to write arcsin(x).

Is the derivative of arcsin(x) the same as the derivative of 1/sin(x)?

No. 1/sin(x) is csc(x), and its derivative is -csc(x) cot(x). The derivative of arcsin(x) is 1/sqrt(1 - x^2).

What is the domain of the derivative of arcsin(x)?

The derivative 1/sqrt(1 - x^2) is defined on the open interval (-1, 1). Arcsin itself is also defined at x = ±1, but the tangent is vertical there.

How do you differentiate arcsin(2x)?

Multiply 1/sqrt(1 - (2x)^2) by the inner derivative 2. The result is 2/sqrt(1 - 4x^2), and it requires |x| < 1/2.

Graph and rules

f(x) f'(x)

Practice More Problems

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Frequently Asked Questions

What is the derivative of asin(x)?▼
The derivative of asin(x) is sqrt(-x^2+1)^(-1).
How do you find the derivative of asin(x)?▼
Apply the standard differentiation rules — the key rule used here is the Inverse Trigonometric Rule. The step-by-step result is sqrt(-x^2+1)^(-1).
Why is the derivative of asin(x) equal to sqrt(-x^2+1)^(-1)?▼
Because differentiation measures how a function changes — applying the rules to asin(x) yields sqrt(-x^2+1)^(-1). See the full step-by-step breakdown above.

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