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Notely Maths 12
Class 12 Maths
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2.5 Q-10

Question Statement

Differentiate the following expressions with respect to xx:

i. cos⁡−1xa\cos^{-1} \frac{x}{a}

ii. 1asin⁡−1(ax)\frac{1}{a} \sin^{-1} \left( \frac{a}{x} \right)

iii. sin⁡−11−x2\sin^{-1} \sqrt{1 - x^{2}}

iv. sec⁡−1(x2+1x2−1)\sec^{-1} \left( \frac{x^2 + 1}{x^2 - 1} \right)

v. cot⁡−1(2x1−x2)\cot^{-1} \left( \frac{2x}{1 - x^2} \right)

vi. cos⁡−1(1−x21+x2)\cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)

Background and Explanation

To differentiate inverse trigonometric functions, we use standard differentiation formulas for each type. Here’s a quick overview:

  1. Inverse Cosine: The derivative of cos⁡−1(u)\cos^{-1}(u) is −11−u2⋅dudx-\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}.
  2. Inverse Sine: The derivative of sin⁡−1(u)\sin^{-1}(u) is 11−u2⋅dudx\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}.
  3. Secant Inverse: The derivative of sec⁡−1(u)\sec^{-1}(u) is 1∣u∣u2−1⋅dudx\frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx}.
  4. Cotangent Inverse: The derivative of cot⁡−1(u)\cot^{-1}(u) is −11+u2⋅dudx-\frac{1}{1 + u^2} \cdot \frac{du}{dx}.

We’ll apply these formulas to each of the given functions.

Solution

i. Differentiate cos⁡−1xa\cos^{-1} \frac{x}{a} with respect to xx

Let y=cos⁡−1xay = \cos^{-1} \frac{x}{a}.

Differentiating both sides:

dydx=−11−(xa)2⋅ddx(xa)\frac{dy}{dx} = -\frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \cdot \frac{d}{dx} \left( \frac{x}{a} \right)

Since ddx(xa)=1a\frac{d}{dx} \left( \frac{x}{a} \right) = \frac{1}{a}, the derivative becomes:

dydx=−11−x2a2⋅1a\frac{dy}{dx} = -\frac{1}{\sqrt{1 - \frac{x^2}{a^2}}} \cdot \frac{1}{a}

Simplifying:

dydx=−1a2−x2⋅1a\frac{dy}{dx} = -\frac{1}{\sqrt{a^2 - x^2}} \cdot \frac{1}{a}

Thus, the final derivative is:

dydx=−1a2+x2\frac{dy}{dx} = -\frac{1}{a^2 + x^2}

ii. Differentiate 1asin⁡−1(ax)\frac{1}{a} \sin^{-1} \left( \frac{a}{x} \right) with respect to xx

Let y=1asin⁡−1(ax)y = \frac{1}{a} \sin^{-1} \left( \frac{a}{x} \right).

Differentiating both sides:

dydx=1a⋅11−(ax)2⋅ddx(ax)\frac{dy}{dx} = \frac{1}{a} \cdot \frac{1}{\sqrt{1 - \left( \frac{a}{x} \right)^2}} \cdot \frac{d}{dx} \left( \frac{a}{x} \right)

Since ddx(ax)=−ax2\frac{d}{dx} \left( \frac{a}{x} \right) = -\frac{a}{x^2}, we get:

dydx=1a⋅1x2−a2⋅(−ax2)\frac{dy}{dx} = \frac{1}{a} \cdot \frac{1}{\sqrt{x^2 - a^2}} \cdot \left( -\frac{a}{x^2} \right)

Simplifying:

dydx=−1xx2−a2\frac{dy}{dx} = \frac{-1}{x \sqrt{x^2 - a^2}}

Thus, the final derivative is:

dydx=−1xx2−a2\frac{dy}{dx} = \frac{-1}{x \sqrt{x^2 - a^2}}

iii. Differentiate sin⁡−11−x2\sin^{-1} \sqrt{1 - x^2} with respect to xx

Let y=sin⁡−11−x2y = \sin^{-1} \sqrt{1 - x^2}.

Differentiating both sides:

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

Simplifying the first term:

dydx=1x2⋅(−2x)\frac{dy}{dx} = \frac{1}{\sqrt{x^2}} \cdot (-2x)

Thus, the derivative becomes:

dydx=−22x1−x2=−11−x2\frac{dy}{dx} = \frac{-2}{2x \sqrt{1 - x^2}} = \frac{-1}{\sqrt{1 - x^2}}

iv. Differentiate sec⁡−1(x2+1x2−1)\sec^{-1} \left( \frac{x^2 + 1}{x^2 - 1} \right) with respect to xx

Let y=sec⁡−1(x2+1x2−1)y = \sec^{-1} \left( \frac{x^2 + 1}{x^2 - 1} \right).

Differentiating both sides:

dydx=1(x2+1x2−1)(x2+1x2−1)2−1⋅ddx(x2+1x2−1)\frac{dy}{dx} = \frac{1}{\left( \frac{x^2 + 1}{x^2 - 1} \right) \sqrt{\left( \frac{x^2 + 1}{x^2 - 1} \right)^2 - 1}} \cdot \frac{d}{dx} \left( \frac{x^2 + 1}{x^2 - 1} \right)

After performing the chain rule and simplification, we obtain:

dydx=−4x(x2+1)x4+2x3+1−x4+2x2−1\frac{dy}{dx} = \frac{-4x}{(x^2 + 1) \sqrt{x^4 + 2x^3 + 1 - x^4 + 2x^2 - 1}}

Simplifying further:

dydx=−4x(x2+1)4x2\frac{dy}{dx} = \frac{-4x}{(x^2 + 1) \sqrt{4x^2}}

Thus, the final derivative is:

dydx=−4x(x2+1)⋅2x=−2x2+1\frac{dy}{dx} = \frac{-4x}{(x^2 + 1) \cdot 2x} = \frac{-2}{x^2 + 1}

v. Differentiate cot⁡−1(2x1−x2)\cot^{-1} \left( \frac{2x}{1 - x^2} \right) with respect to xx

Let y=cot⁡−1(2x1−x2)y = \cot^{-1} \left( \frac{2x}{1 - x^2} \right).

Differentiating both sides:

dydx=−11+(2x1−x2)2⋅ddx(2x1−x2)\frac{dy}{dx} = -\frac{1}{1 + \left( \frac{2x}{1 - x^2} \right)^2} \cdot \frac{d}{dx} \left( \frac{2x}{1 - x^2} \right)

After applying the quotient rule and simplifying, we get:

dydx=−21+x2\frac{dy}{dx} = \frac{-2}{1 + x^2}

vi. Differentiate cos⁡−1(1−x21+x2)\cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) with respect to xx

Let y=cos⁡−1(1−x21+x2)y = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right).

Differentiating both sides:

dydx=−11−(1−x21+x2)2⋅ddx(1−x21+x2)\frac{dy}{dx} = \frac{-1}{\sqrt{1 - \left( \frac{1 - x^2}{1 + x^2} \right)^2}} \cdot \frac{d}{dx} \left( \frac{1 - x^2}{1 + x^2} \right)

After applying the quotient rule and simplifying, we get:

dydx=2x2+1\frac{dy}{dx} = \frac{2}{x^2 + 1}

Key Formulas or Methods Used

  1. Derivative of Inverse Cosine: ddxcos⁡−1(u)=−11−u2⋅dudx\frac{d}{dx} \cos^{-1}(u) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}

  2. Derivative of Inverse Sine: ddxsin⁡−1(u)=11−u2⋅dudx\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}

  3. Derivative of Secant Inverse: ddxsec⁡−1(u)=1∣u∣u2−1⋅dudx\frac{d}{dx} \sec^{-1}(u) = \frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx}

  4. Derivative of Cotangent Inverse: ddxcot⁡−1(u)=−11+u2⋅dudx\frac{d}{dx} \cot^{-1}(u) = -\frac{1}{1 + u^2} \cdot \frac{du}{dx}