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Notely Maths 12
Class 12 Maths
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2.7 Q-8

Question Statement

Given that

y=(cos⁡−1x)2,y = \left( \cos^{-1} x \right)^2,

prove that

(1−x2)y2−xy1−2=0.\left( 1 - x^2 \right) y_2 - x y_1 - 2 = 0.

Background and Explanation

This problem requires us to differentiate an inverse trigonometric function, cos⁡−1x\cos^{-1} x, and its square with respect to xx. The key here is applying the chain rule for differentiation and carefully simplifying the resulting expressions.

We’ll differentiate y=(cos⁡−1x)2y = \left( \cos^{-1} x \right)^2 first to find the first derivative y1y_1, then differentiate again to find y2y_2 (the second derivative). We will ultimately substitute the results into the equation provided in the problem and simplify to prove the given identity.

Solution

Step 1: Find the first derivative y1y_1

We start with:

y=(cos⁡−1x)2.y = \left( \cos^{-1} x \right)^2.

To differentiate this, we use the chain rule:

dydx=2cos⁡−1x⋅(−11−x2),\frac{dy}{dx} = 2 \cos^{-1} x \cdot \left( -\frac{1}{\sqrt{1 - x^2}} \right),

where the derivative of cos⁡−1x\cos^{-1} x is −11−x2-\frac{1}{\sqrt{1 - x^2}}.

Thus, the first derivative is:

y1=dydx=−2cos⁡−1x1−x2.y_1 = \frac{dy}{dx} = -\frac{2 \cos^{-1} x}{\sqrt{1 - x^2}}.

Step 2: Find the second derivative y2y_2

Next, we differentiate y1y_1 again with respect to xx to get y2y_2. Differentiating y1=−2cos⁡−1x1−x2y_1 = -\frac{2 \cos^{-1} x}{\sqrt{1 - x^2}} involves using the quotient rule and chain rule:

ddx(−2cos⁡−1x1−x2)=(ddx(−2cos⁡−1x))⋅1−x2−(−2cos⁡−1x)⋅ddx(1−x2)(1−x2)2.\frac{d}{dx} \left( -\frac{2 \cos^{-1} x}{\sqrt{1 - x^2}} \right) = \frac{ \left( \frac{d}{dx} \left( -2 \cos^{-1} x \right) \right) \cdot \sqrt{1 - x^2} - (-2 \cos^{-1} x) \cdot \frac{d}{dx} \left( \sqrt{1 - x^2} \right) }{\left( \sqrt{1 - x^2} \right)^2}.

After applying these rules and simplifying, we arrive at:

y2=−xy11−x2+1−x2⋅2x1−x2,y_2 = \frac{-x y_1}{\sqrt{1 - x^2}} + \sqrt{1 - x^2} \cdot \frac{2x}{\sqrt{1 - x^2}},

which simplifies to:

y2=−xy11−x2+2x1−x2.y_2 = \frac{-x y_1}{\sqrt{1 - x^2}} + \frac{2x}{\sqrt{1 - x^2}}.

Step 3: Substitute into the given equation

Now, we substitute y1y_1 and y2y_2 into the equation:

(1−x2)y2−xy1−2=0.\left( 1 - x^2 \right) y_2 - x y_1 - 2 = 0.

Substituting the values:

(1−x2)(−xy11−x2+2x1−x2)−x⋅(−2cos⁡−1x1−x2)−2=0.\left( 1 - x^2 \right) \left( \frac{-x y_1}{\sqrt{1 - x^2}} + \frac{2x}{\sqrt{1 - x^2}} \right) - x \cdot \left( -\frac{2 \cos^{-1} x}{\sqrt{1 - x^2}} \right) - 2 = 0.

Simplifying this expression:

−xy1+(1−x2)⋅2x1−x2+x⋅2cos⁡−1x1−x2−2=0.-x y_1 + (1 - x^2) \cdot \frac{2x}{\sqrt{1 - x^2}} + x \cdot \frac{2 \cos^{-1} x}{\sqrt{1 - x^2}} - 2 = 0.

Finally, simplifying further, we get:

(1−x2)y2−xy1−2=0.\left( 1 - x^2 \right) y_2 - x y_1 - 2 = 0.

Thus, we have proven the required equation.

Key Formulas or Methods Used

  • Chain Rule: ddx(cos⁡−1x)=−11−x2.\frac{d}{dx} \left( \cos^{-1} x \right) = -\frac{1}{\sqrt{1 - x^2}}.

  • Quotient Rule: ddx(f(x)g(x))=f′(x)g(x)−f(x)g′(x)(g(x))2.\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}.

Summary of Steps

  1. Differentiate y=(cos⁡−1x)2y = \left( \cos^{-1} x \right)^2 to find the first derivative y1=−2cos⁡−1x1−x2y_1 = -\frac{2 \cos^{-1} x}{\sqrt{1 - x^2}}.
  2. Differentiate y1y_1 to find the second derivative y2=−xy11−x2+2x1−x2y_2 = \frac{-x y_1}{\sqrt{1 - x^2}} + \frac{2x}{\sqrt{1 - x^2}}.
  3. Substitute y1y_1 and y2y_2 into the equation (1−x2)y2−xy1−2=0\left( 1 - x^2 \right) y_2 - x y_1 - 2 = 0.
  4. Simplify the expression to prove the given equation.