#### Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 26

$\frac{d y}{d x}=t$

Hint:

Use quotient rule and   $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

Given:

\begin{aligned} &x=\frac{1+\log t}{t^{2}} \\ &y=\frac{3+2 \log t}{t} \end{aligned}

Solution:

$x=\frac{1+\log t}{t^{2}} \\$

$\frac{d x}{d t}=\frac{d\left(\frac{1+\log t}{t^{2}}\right)}{d t} \\$

\begin{aligned} &=\frac{t^{2} \frac{d(1+\log t)}{d t}-(1+\log t) \frac{d t^{2}}{d t}}{\left(t^{2}\right)^{2}} \end{aligned}                                                                            [Using quotient rule]

$\frac{d x}{d t}=\frac{t\left[\frac{d(1)}{d t}+\frac{d \log t}{d t}\right]-(1+\log t) \cdot 2 t}{t^{4}} \\$

$=\frac{t^{2}\left(0+\frac{1}{t}\right)-2 t-2 t \log t}{t^{4}} \\$

$=\frac{t-2 t-2 t \log t}{t^{4}} \\$

\begin{aligned} & &=\frac{-t-2 t \log t}{t^{4}} \end{aligned}

$\frac{d x}{d t}=\frac{-(1+2 \log t)}{t^{3}} \\$                                                                                                                                        (1)

\begin{aligned} & &y=\frac{3+2 \log t}{t} \end{aligned}

$\frac{d y}{d t}=\frac{d\left(\frac{3+2 \log t}{t}\right)}{d t} \\$

\begin{aligned} &=\frac{t \cdot \frac{d(3+2 \log t)}{d t}-(3+2 \log t) \frac{d t}{d t}}{t^{2}} \end{aligned}                                                                      [Using quotient rule]

$=\frac{t\left(\frac{2}{t}\right)-(3+2 \log t)}{t^{2}} \\$

$\frac{d y}{d t}=\frac{2-3-2 \log t}{t^{2}} \\$

$=\frac{-1-2 \log t}{t^{2}} \\$

\begin{aligned} & &\frac{d y}{d t}=\frac{-(1+2 \log t)}{t^{2}} \end{aligned}                                                                                                                                        (2)

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

So put $\frac{d x}{d t} \text { and } \frac{d y}{d t}$   from equation (1) and (2) respectively

$\frac{d y}{d x}=\frac{\frac{-(1+2 \log t)}{t^{2}}}{\frac{-(1+2 \log t)}{t^{3}}}=\frac{-(1+2 \log t) \times t^{3}}{-(1+2 \log t) \times t^{2}}$

$\\ =\frac{t^{3}}{t^{2}}=t \\$

\begin{aligned} & &\frac{d y}{d x}=t \end{aligned}