Question

If \[ y=\left(\frac{a+x}{b+x}\right)^{a+b+2x}, \] then \[ \left.\frac{dy}{dx}\right|_{x=0} \] is

• A. \(1\)
• B. \(\log\frac{a}{b}\)
• C. \[ \left( 2\log\frac{a}{b} +\frac{b^{2}-a^{2}}{ab} \right) \left(\frac{a}{b}\right)^{a+b} \]
• D. \[ \left( \log\frac{a}{b} +\frac{ab}{\,b-a\,} \right) \left(\frac{b}{a}\right)^{a+b} \]

Hint:

Whenever both the base and exponent contain variables, \[ y=u(x)^{v(x)}, \] apply logarithmic differentiation: \[ \log y=v\log u. \]

Answer 1

The Correct Option is C

Concept: For functions of the form \[ y=u(x)^{v(x)}, \] use logarithmic differentiation: \[ \log y=v\log u. \]

Step 1: Take logarithm on both sides. \[ \log y = (a+b+2x) \log\left(\frac{a+x}{b+x}\right) \] Differentiating, \[\begin{aligned} \frac1y\frac{dy}{dx} &= 2\log\left(\frac{a+x}{b+x}\right) +(a+b+2x) \frac{d}{dx} \log\left(\frac{a+x}{b+x}\right) \end{aligned}\]

Step 2: Differentiate the logarithmic term. \[\begin{aligned} \frac{d}{dx} \log\left(\frac{a+x}{b+x}\right) &= \frac1{a+x}-\frac1{b+x} \end{aligned}\] Hence, \[\begin{aligned} \frac1y\frac{dy}{dx} = 2\log\left(\frac{a+x}{b+x}\right) + (a+b+2x) \left( \frac1{a+x}-\frac1{b+x} \right) \end{aligned}\]

Step 3: Put \(x=0\). \[ y(0) = \left(\frac ab\right)^{a+b} \] and \[\begin{aligned} \left.\frac1y\frac{dy}{dx}\right|_{x=0} &= 2\log\frac ab + (a+b) \left( \frac1a-\frac1b \right) \end{aligned}\] \[\begin{aligned} &= 2\log\frac ab + (a+b)\frac{b-a}{ab} \end{aligned}\] \[\begin{aligned} &= 2\log\frac ab + \frac{b^2-a^2}{ab} \end{aligned}\]

Step 4: Find \(\left.\dfrac{dy}{dx}\right|_{x=0}\). \[\begin{aligned} \left.\frac{dy}{dx}\right|_{x=0} &= \left(\frac ab\right)^{a+b} \left( 2\log\frac ab + \frac{b^2-a^2}{ab} \right) \end{aligned}\] \[\begin{aligned} \boxed{ \left( 2\log\frac ab +\frac{b^2-a^2}{ab} \right) \left(\frac ab\right)^{a+b} } \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.