How To Differentiate Sin⁻¹(X) Using The Chain Rule | Math Tutorial And Example Problem

ddx(sin−1x)∣∣x=12=

Correct. If y=sin−1xy=sin⁡−1x, then siny=xsin⁡y=x. Using the chain rule to differentiate this implicit relation gives cosydydx=1⇒dydx=1cosycos⁡ydydx=1⇒dydx=1cos⁡y. Since y=sin−1xy=sin⁡−1x implies that −π2≤y≤π2−π2≤y≤π2, the value of cosycos⁡y is nonnegative. Then cos2y=1−sin2y=1−x2⇒cosy=1−x2−−−−−√cos⁡2y=1−sin⁡2y=1−x2⇒cos⁡y=1−x2. Therefore, dydx=1cosy=11−x2√dydx=1cos⁡y=11−x2. This is evaluated at x=12x=12.

To differentiate `sin⁻¹(x)`, we need to use the chain rule.

Let `y = sin⁻¹(x)`, then `sin(y) = x`.

Differentiating implicitly with respect to `x`, we get:

`cos(y) * dy/dx = 1`

Solving for `dy/dx`:

`dy/dx = 1/cos(y)`

We can substitute `y = sin⁻¹(x)` to get:

`dy/dx = 1/cos(sin⁻¹(x))`

Now, using the Pythagorean identity, `cos²θ + sin²θ = 1`, we can solve for `cos(θ)` in terms of `sin(θ)`:

`cos(θ) = √(1 – sin²(θ))`

Substituting `θ = sin⁻¹(x)` gives:

`cos(sin⁻¹(x)) = √(1 – x²)`

Therefore:

`dy/dx = 1/√(1 – x²)`

Differentiating with respect to `x`, we get:

`d²y/dx² = d/dx(1/√(1 – x²))`

Using the chain rule:

`d²y/dx² = -1/(1 – x²)^(3/2) * d/dx(x²)`

`d²y/dx² = -1/(1 – x²)^(3/2) * 2x`

Evaluating at `x = 1/2`:

`d²y/dx²∣∣x=1/2 = -1/(1 – (1/2)²)^(3/2) * 2(1/2)`

`d²y/dx²∣∣x=1/2 = -4/3√3`

Therefore:

`ddx(sin⁻¹(x))∣∣x=1/2 = -4/3√3`

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