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Math  /  Calculus

Question(1 point)
Differentiate y=1−x2sin⁡−1xy=\sqrt{1-x^{2}} \sin ^{-1} x y′=y^{\prime}=

Studdy Solution

STEP 1

What is this asking? Find the derivative of a function that's the product of a square root and an inverse sine function. Watch out! Don't forget the chain rule and the product rule!
Also, remember the derivative of inverse sine.

STEP 2

1. Rewrite the function
2. Apply the product rule
3. Calculate the derivatives of the individual functions
4. Simplify the result

STEP 3

Let's **rewrite** the function yy using exponents to make it easier to differentiate: y=(1−x2)12⋅sin⁡−1(x)y = (1 - x^2)^{\frac{1}{2}} \cdot \sin^{-1}(x) This way, we can clearly see the power we'll need to deal with when we differentiate.

STEP 4

Now, let's **apply** the product rule, which states: If y=u⋅vy = u \cdot v, then y′=u′⋅v+u⋅v′y' = u' \cdot v + u \cdot v'. In our case, u=(1−x2)12u = (1 - x^2)^{\frac{1}{2}} and v=sin⁡−1(x)v = \sin^{-1}(x).

STEP 5

First, let's **find** u′u', the derivative of u=(1−x2)12u = (1 - x^2)^{\frac{1}{2}}, using the chain rule: u′=12(1−x2)−12⋅(−2x)=−x(1−x2)−12=−x1−x2u' = \frac{1}{2}(1 - x^2)^{-\frac{1}{2}} \cdot (-2x) = -x(1 - x^2)^{-\frac{1}{2}} = \frac{-x}{\sqrt{1 - x^2}} Remember, the chain rule says to take the derivative of the outside function, multiplied by the derivative of the inside function.

STEP 6

Next, let's **find** v′v', the derivative of v=sin⁡−1(x)v = \sin^{-1}(x).
Remember, the derivative of sin⁡−1(x)\sin^{-1}(x) is 11−x2\frac{1}{\sqrt{1 - x^2}}: v′=11−x2v' = \frac{1}{\sqrt{1 - x^2}}

STEP 7

Now, let's **put it all together** using the product rule formula from earlier: y′=u′v+uv′y' = u'v + uv' y′=−x1−x2⋅sin⁡−1(x)+(1−x2)12⋅11−x2y' = \frac{-x}{\sqrt{1 - x^2}} \cdot \sin^{-1}(x) + (1 - x^2)^{\frac{1}{2}} \cdot \frac{1}{\sqrt{1 - x^2}}y′=−xsin⁡−1(x)1−x2+1−x21−x2y' = \frac{-x \sin^{-1}(x)}{\sqrt{1 - x^2}} + \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^2}}y′=−xsin⁡−1(x)1−x2+1y' = \frac{-x \sin^{-1}(x)}{\sqrt{1 - x^2}} + 1

STEP 8

y′=1−xsin⁡−1(x)1−x2y' = 1 - \frac{x \sin^{-1}(x)}{\sqrt{1 - x^2}}

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