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Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 20

Answers (1)

Answer:

(a)

Hint:

We must know about the derivative of logarithm.

Given:

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

Explanation:

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

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

$\begin{aligned} &\ &\frac{y}{x}=\log x-\log (a+b x) \end{aligned}$

Differentiate with respect to $x$

$\frac{x \frac{d y}{d x}-y}{x^{2}}=\frac{1}{x}-\frac{b}{a+b x} \$

$\begin{aligned} &\ &x \frac{d y}{d x}-y=x-\frac{b x^{2}}{a+b x} \end{aligned}$

Differentiate with respect to $x$

$x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-\frac{d y}{d x}=1-\left(\frac{(a+b x) 2 b x-b x^{2}(b)}{(a+b x)^{2}}\right) \\$

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

$\begin{aligned} & &x \frac{d^{2} y}{d x^{2}}=\left(\frac{(a+b x)[a+b x-2 b x]+b^{2} x^{2}}{(a+b x)^{2}}\right) \end{aligned}$

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

$\begin{aligned} &\\ &x \frac{d^{2} y}{d x^{2}}=\left(\frac{a^{2}-b^{2} x^{2}+b^{2} x^{2}}{(a+b x)^{2}}\right) \end{aligned}$

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

$\\ x \frac{d^{2} y}{d x^{2}}=\left(\frac{a}{a+b x}\right)^{2} \$ … (i)

And $\begin{aligned} \ &y=x \log \left(\frac{x}{a+b x}\right) \end{aligned}$

Differentiate with respect to $x$

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

$\frac{d y}{d x}=\frac{a}{a+b x}+\frac{y}{x}$

$\begin{aligned} \\ &\frac{d y}{d x}-\frac{y}{x}=\frac{a}{a+b x} \end{aligned}$

From (i)

$x \frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}-\frac{y}{x}\right)^{2}$

$\begin{aligned} & \\ &x^{2} \frac{d^{2} y}{d x^{2}}=\frac{1}{x^{2}}\left(x \frac{d y}{d x}-y\right)^{2} \end{aligned}$

$x^{3} \frac{d^{2} y}{d x^{2}}=\left(x \frac{d y}{d x}-y\right)^{2}$

$\begin{aligned} &\\ &x^{3} y_{2}=\left(x y_{1}-y\right)^{2} \end{aligned}$

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