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2.3 Q-3

Question Statement

Differentiate the following expression with respect to xx: a+xaβˆ’x\frac{a + x}{a - x}


Background and Explanation

To solve this problem, we will use the quotient rule for differentiation. The quotient rule is applied when you are differentiating a function that is the ratio of two functions, such as u(x)v(x)\frac{u(x)}{v(x)}. The quotient rule states: ddx(u(x)v(x))=v(x)β‹…uβ€²(x)βˆ’u(x)β‹…vβ€²(x)(v(x))2\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}

Here, u(x)=a+xu(x) = a + x and v(x)=aβˆ’xv(x) = a - x, so we need to differentiate both the numerator and the denominator.


Solution

Let the function be: Y=a+xaβˆ’xY = \frac{a + x}{a - x}

Now, differentiate both sides with respect to xx using the quotient rule:

  1. Differentiate the numerator (u(x)=a+xu(x) = a + x): uβ€²(x)=1u'(x) = 1

  2. Differentiate the denominator (v(x)=aβˆ’xv(x) = a - x): vβ€²(x)=βˆ’1v'(x) = -1

  3. Apply the quotient rule: Using the formula for the quotient rule: dydx=v(x)β‹…uβ€²(x)βˆ’u(x)β‹…vβ€²(x)(v(x))2\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}

    Substitute the values for u(x)u(x), v(x)v(x), uβ€²(x)u'(x), and vβ€²(x)v'(x): dydx=(aβˆ’x)β‹…1βˆ’(a+x)β‹…(βˆ’1)(aβˆ’x)2\frac{dy}{dx} = \frac{(a - x) \cdot 1 - (a + x) \cdot (-1)}{(a - x)^2}

  4. Simplify the expression: dydx=(aβˆ’x)+(a+x)(aβˆ’x)2\frac{dy}{dx} = \frac{(a - x) + (a + x)}{(a - x)^2}

    Combine like terms in the numerator: dydx=2a(aβˆ’x)2\frac{dy}{dx} = \frac{2a}{(a - x)^2}

Thus, the derivative of the given function is: dydx=2a(aβˆ’x)2\frac{dy}{dx} = \frac{2a}{(a - x)^2}

Final Answer: dydx=2a(aβˆ’x)2\frac{dy}{dx} = \frac{2a}{(a - x)^2}


Key Formulas or Methods Used

  • Quotient Rule:
    If you have a function in the form u(x)v(x)\frac{u(x)}{v(x)}, the derivative is: ddx(u(x)v(x))=v(x)β‹…uβ€²(x)βˆ’u(x)β‹…vβ€²(x)(v(x))2\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}

Summary of Steps

  1. Identify the function:
    Y=a+xaβˆ’xY = \frac{a + x}{a - x}

  2. Apply the quotient rule:

    • Differentiate the numerator u(x)=a+xu(x) = a + x to get uβ€²(x)=1u'(x) = 1
    • Differentiate the denominator v(x)=aβˆ’xv(x) = a - x to get vβ€²(x)=βˆ’1v'(x) = -1
  3. Substitute into the quotient rule formula: dydx=(aβˆ’x)β‹…1βˆ’(a+x)β‹…(βˆ’1)(aβˆ’x)2\frac{dy}{dx} = \frac{(a - x) \cdot 1 - (a + x) \cdot (-1)}{(a - x)^2}

  4. Simplify the expression: dydx=2a(aβˆ’x)2\frac{dy}{dx} = \frac{2a}{(a - x)^2}

  5. Final Answer: dydx=2a(aβˆ’x)2\frac{dy}{dx} = \frac{2a}{(a - x)^2}