If y = sin^-1 [x√(1-x) - √x√(1-x^2)] and 0 < x < 1, then find dy/dx.
The given problem is:
y = sin − 1 [ x 1 − x − x 1 − x 2 ] , 0 < x < 1 y = \sin^{-1} \left[ x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^2} \right], \, 0 < x < 1 y = sin − 1 [ x 1 − x − x 1 − x 2 ] , 0 < x < 1
We are tasked with finding d y d x \frac{dy}{dx} d x d y
Step-by-step solution:
Simplify the argument of sin − 1 \sin^{-1} sin − 1
Let:
z = x 1 − x − x 1 − x 2 . z = x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^2}. z = x 1 − x − x 1 − x 2 .
Differentiate y y y x x x
From the formula for the derivative of sin − 1 ( u ) \sin^{-1}(u) sin − 1 ( u ) d y d x = 1 1 − z 2 ⋅ d z d x . \frac{dy}{dx} = \frac{1}{\sqrt{1 - z^2}} \cdot \frac{dz}{dx}. d x d y = 1 − z 2 1 ⋅ d x d z .
So, we need to calculate z 2 z^2 z 2 d z d x \frac{dz}{dx} d x d z
Calculation of z 2 z^2 z 2
z = x 1 − x − x 1 − x 2 . z = x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^2}. z = x 1 − x − x 1 − x 2 .
Square z z z z 2 = ( x 1 − x − x 1 − x 2 ) 2 . z^2 = \left(x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^2}\right)^2. z 2 = ( x 1 − x − x 1 − x 2 ) 2 .
Expanding this:
z 2 = ( x 1 − x ) 2 + ( x 1 − x 2 ) 2 − 2 ( x 1 − x ) ( x 1 − x 2 ) . z^2 = \left(x \sqrt{1 - x}\right)^2 + \left(\sqrt{x} \sqrt{1 - x^2}\right)^2 - 2\left(x \sqrt{1 - x}\right)\left(\sqrt{x} \sqrt{1 - x^2}\right). z 2 = ( x 1 − x ) 2 + ( x 1 − x 2 ) 2 − 2 ( x 1 − x ) ( x 1 − x 2 ) .
Compute each term:
First term:
( x 1 − x ) 2 = x 2 ( 1 − x ) . \left(x \sqrt{1 - x}\right)^2 = x^2 (1 - x). ( x 1 − x ) 2 = x 2 ( 1 − x ) .
Second term:
( x 1 − x 2 ) 2 = x ( 1 − x 2 ) . \left(\sqrt{x} \sqrt{1 - x^2}\right)^2 = x (1 - x^2). ( x 1 − x 2 ) 2 = x ( 1 − x 2 ) .
Third term:
2 ( x 1 − x ) ( x 1 − x 2 ) = 2 x x ( 1 − x ) ( 1 − x 2 ) . 2 \left(x \sqrt{1 - x}\right)\left(\sqrt{x} \sqrt{1 - x^2}\right) = 2x \sqrt{x(1 - x)(1 - x^2)}. 2 ( x 1 − x ) ( x 1 − x 2 ) = 2 x x ( 1 − x ) ( 1 − x 2 ) .
Combine the terms:
z 2 = x 2 ( 1 − x ) + x ( 1 − x 2 ) − 2 x x ( 1 − x ) ( 1 − x 2 ) . z^2 = x^2 (1 - x) + x (1 - x^2) - 2x \sqrt{x(1 - x)(1 - x^2)}. z 2 = x 2 ( 1 − x ) + x ( 1 − x 2 ) − 2 x x ( 1 − x ) ( 1 − x 2 ) .
Simplify:
z 2 = x 2 − x 3 + x − x 3 − 2 x x ( 1 − x ) ( 1 − x 2 ) . z^2 = x^2 - x^3 + x - x^3 - 2x \sqrt{x(1 - x)(1 - x^2)}. z 2 = x 2 − x 3 + x − x 3 − 2 x x ( 1 − x ) ( 1 − x 2 ) . z 2 = x − 2 x 3 − 2 x x ( 1 − x ) ( 1 − x 2 ) . z^2 = x - 2x^3 - 2x \sqrt{x(1 - x)(1 - x^2)}. z 2 = x − 2 x 3 − 2 x x ( 1 − x ) ( 1 − x 2 ) .
Calculation of d z d x \frac{dz}{dx} d x d z
Differentiate z = x 1 − x − x 1 − x 2 z = x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^2} z = x 1 − x − x 1 − x 2
First term:
d d x ( x 1 − x ) = 1 − x + x ⋅ − 1 2 1 − x = 1 − x − x 2 1 − x . \frac{d}{dx}\left(x \sqrt{1 - x}\right) = \sqrt{1 - x} + x \cdot \frac{-1}{2\sqrt{1 - x}} = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}. d x d ( x 1 − x ) = 1 − x + x ⋅ 2 1 − x − 1 = 1 − x − 2 1 − x x .
Second term:
d d x ( x 1 − x 2 ) = 1 2 x ⋅ 1 − x 2 + x ⋅ − 2 x 2 1 − x 2 . \frac{d}{dx}\left(\sqrt{x} \sqrt{1 - x^2}\right) = \frac{1}{2\sqrt{x}} \cdot \sqrt{1 - x^2} + \sqrt{x} \cdot \frac{-2x}{2\sqrt{1 - x^2}}. d x d ( x 1 − x 2 ) = 2 x 1 ⋅ 1 − x 2 + x ⋅ 2 1 − x 2 − 2 x . = 1 − x 2 2 x − x 1 − x 2 ⋅ x . = \frac{\sqrt{1 - x^2}}{2\sqrt{x}} - \frac{x}{\sqrt{1 - x^2}} \cdot \sqrt{x}. = 2 x 1 − x 2 − 1 − x 2 x ⋅ x .
Combine both terms for d z d x \frac{dz}{dx} d x d z d z d x = ( 1 − x − x 2 1 − x ) − ( 1 − x 2 2 x − x x 1 − x 2 ) . \frac{dz}{dx} = \left(\sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}\right) - \left(\frac{\sqrt{1 - x^2}}{2\sqrt{x}} - \frac{x \sqrt{x}}{\sqrt{1 - x^2}}\right). d x d z = ( 1 − x − 2 1 − x x ) − ( 2 x 1 − x 2 − 1 − x 2 x x ) .
Final Answer:
Substitute d z d x \frac{dz}{dx} d x d z z 2 z^2 z 2 d y d x = 1 1 − z 2 ⋅ d z d x . \frac{dy}{dx} = \frac{1}{\sqrt{1 - z^2}} \cdot \frac{dz}{dx}. d x d y = 1 − z 2 1 ⋅ d x d z .
This completes the derivative calculation.
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Tip:
For complex derivatives involving square roots, breaking down the problem into smaller parts and simplifying step-by-step is crucial to avoid mistakes.