Question

If $f\left(x\right)={\left(\frac{a+x}{b+x}\right)}^{a+b+2x}$ then, $f\text{'}\left(0\right)=$

• A. $2\log \frac{a}{b}+\frac{b^2-a^2}{ab}$
• B. ${\left(\frac{a}{b}\right)}^{a+b}\left[2\log \frac{a}{b}+\frac{b^2-a^2}{ab}\right]$
• C. ${\left(\frac{a}{b}\right)}^{a+b}\left[\frac{b^2-a^2}{ab}\right]$
• D. None of these

Answer 1

The correct option is B

${\left(\frac{a}{b}\right)}^{a+b}\left[2\log \frac{a}{b}+\frac{b^2-a^2}{ab}\right]$

Explanation for the correct option.

Step 1: Find the first derivative.

Let, $y=f\left(x\right)$, so $y={\left(\frac{a+x}{b+x}\right)}^{a+b+2x}$.

Taking log on both sides we get,

$\log y=\left(a+b+2x\right)\log \left(\frac{a+x}{b+x}\right)$

By differentiating with respect to $x$ we get,

$\begin{array}{rcl}\frac{1}{y}\frac{dy}{dx}& =& 2\log \left(\frac{a+x}{b+x}\right)+\left[\left(a+b+2x\right)\times \frac{1}{\left({\displaystyle \frac{a+x}{b+x}}\right)}\times \frac{\left[\left(b+x\right)\left(1\right)-\left(a+x\right)\left(1\right)\right]}{{\left(b+x\right)}^2}\right]\\ & \Rightarrow & \frac{1}{y}\frac{dy}{dx}=2\log \left(\frac{a+x}{b+x}\right)+\left[\left(a+b+2x\right)\times \frac{b+x}{a+x}\times \frac{b-a}{{\left(b+x\right)}^2}\right]\\ & \Rightarrow & \frac{1}{y}\frac{dy}{dx}=2\log \left(\frac{a+x}{b+x}\right)+\left[\frac{\left(a+b+2x\right)\left(b-a\right)}{\left(a+x\right)\left(b+x\right)}\right]\\ \Rightarrow \frac{dy}{dx}& =& y\left[2\log \left(\frac{a+x}{b+x}\right)+\left\{\frac{\left(a+b+2x\right)\left(b-a\right)}{\left(a+x\right)\left(b+x\right)}\right\}\right]...\left(1\right)\end{array}$

Step 2: Find $f\text{'}\left(0\right)$.

Now, $\frac{dy}{dx}=f\text{'}\left(x\right)$ and $y=f\left(x\right)$. So, $\left(1\right)$ can be written as

$\begin{array}{rcl}f\text{'}\left(x\right)& =& {\left(\frac{a+x}{b+x}\right)}^{a+b+2x}\left[2\log \left(\frac{a+x}{b+x}\right)+\left\{\frac{\left(a+b+2x\right)\left(b-a\right)}{\left(a+x\right)\left(b+x\right)}\right\}\right]\end{array}$

By putting $x=0$, we get

$\begin{array}{rcl}f\text{'}\left(0\right)& =& {\left(\frac{a}{b}\right)}^{a+b}\left[2\log \left(\frac{a}{b}\right)+\left\{\frac{\left(a+b\right)\left(b-a\right)}{\left(a\right)\left(b\right)}\right\}\right]\\ & =& {\left(\frac{a}{b}\right)}^{a+b}\left[2\log \frac{a}{b}+\frac{b^2-a^2}{ab}\right]\end{array}$

Hence, option B is correct.