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Notely Maths 12
Class 12 Maths
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2.3 Q-12

Question Statement

Differentiate the following function with respect to xx:
y=a−xa+xy = \sqrt{\frac{a - x}{a + x}}

Background and Explanation

This problem involves differentiating a function that combines square roots and fractions. To solve it, we need to:

  1. Use the chain rule for the square root function.
  2. Apply the quotient rule to differentiate the fraction inside the square root.

Key Rules:

  1. Chain Rule:
    If f(x)=g(x)f(x) = \sqrt{g(x)}, then:
    ddxg(x)=12g(x)⋅dg(x)dx\frac{d}{dx} \sqrt{g(x)} = \frac{1}{2\sqrt{g(x)}} \cdot \frac{dg(x)}{dx}
  2. Quotient Rule:
    If f(x)=uvf(x) = \frac{u}{v}, then:
    ddx(uv)=v⋅dudx−u⋅dvdxv2\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

Solution

Step 1: Write the function in terms of powers

We can rewrite the given function as: y=(a−xa+x)1/2y = \left(\frac{a - x}{a + x}\right)^{1/2}

Step 2: Differentiate using the chain rule

Using the chain rule for square roots: dydx=12a−xa+x⋅ddx(a−xa+x)\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{a - x}{a + x}}} \cdot \frac{d}{dx} \left(\frac{a - x}{a + x}\right)

Step 3: Differentiate the fraction using the quotient rule

Let:

  • u=a−xu = a - x, so dudx=−1\frac{du}{dx} = -1
  • v=a+xv = a + x, so dvdx=1\frac{dv}{dx} = 1

Using the quotient rule: ddx(a−xa+x)=(a+x)⋅ddx(a−x)−(a−x)⋅ddx(a+x)(a+x)2\frac{d}{dx}\left(\frac{a - x}{a + x}\right) = \frac{(a + x) \cdot \frac{d}{dx}(a - x) - (a - x) \cdot \frac{d}{dx}(a + x)}{(a + x)^2}
Substitute the derivatives: ddx(a−xa+x)=(a+x)(−1)−(a−x)(1)(a+x)2\frac{d}{dx}\left(\frac{a - x}{a + x}\right) = \frac{(a + x)(-1) - (a - x)(1)}{(a + x)^2}
Simplify the numerator: ddx(a−xa+x)=−(a+x)−(a−x)(a+x)2\frac{d}{dx}\left(\frac{a - x}{a + x}\right) = \frac{-(a + x) - (a - x)}{(a + x)^2}
ddx(a−xa+x)=−a−x−a+x(a+x)2\frac{d}{dx}\left(\frac{a - x}{a + x}\right) = \frac{-a - x - a + x}{(a + x)^2}
ddx(a−xa+x)=−2a(a+x)2\frac{d}{dx}\left(\frac{a - x}{a + x}\right) = \frac{-2a}{(a + x)^2}

Step 4: Substitute back into the chain rule

Now substitute into the chain rule formula: dydx=12a−xa+x⋅−2a(a+x)2\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{a - x}{a + x}}} \cdot \frac{-2a}{(a + x)^2}

Step 5: Simplify

Simplify the expression:

  1. Rewrite a−xa+x\sqrt{\frac{a - x}{a + x}} as a−xa+x\frac{\sqrt{a - x}}{\sqrt{a + x}}: dydx=12⋅a−xa+x⋅−2a(a+x)2\frac{dy}{dx} = \frac{1}{2 \cdot \frac{\sqrt{a - x}}{\sqrt{a + x}}} \cdot \frac{-2a}{(a + x)^2}
  2. Invert the denominator: dydx=a+x2a−x⋅−2a(a+x)2\frac{dy}{dx} = \frac{\sqrt{a + x}}{2\sqrt{a - x}} \cdot \frac{-2a}{(a + x)^2}
  3. Simplify: dydx=−aa+xa−x⋅(a+x)3/2\frac{dy}{dx} = \frac{-a \sqrt{a + x}}{\sqrt{a - x} \cdot (a + x)^{3/2}}

Thus, the derivative is: dydx=−a(a−x)(a+x)3\frac{dy}{dx} = \frac{-a}{\sqrt{(a - x)(a + x)^3}}

Key Formulas or Methods Used

  1. Chain Rule for Square Roots:
    ddxg(x)=12g(x)⋅dg(x)dx\frac{d}{dx}\sqrt{g(x)} = \frac{1}{2\sqrt{g(x)}} \cdot \frac{dg(x)}{dx}
  2. Quotient Rule:
    ddx(uv)=v⋅dudx−u⋅dvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

Summary of Steps

  1. Rewrite the function as y=(a−xa+x)1/2y = \left(\frac{a - x}{a + x}\right)^{1/2}.
  2. Use the chain rule to differentiate the square root.
  3. Apply the quotient rule to find the derivative of a−xa+x\frac{a - x}{a + x}.
  4. Simplify the result step-by-step to obtain: dydx=−a(a−x)(a+x)3\frac{dy}{dx} = \frac{-a}{\sqrt{(a - x)(a + x)^3}}