#### Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 20

$\frac{d y}{d x}=\frac{(a)^{t+\frac{1}{t}} \cdot \log a}{a\left(t+\frac{1}{t}\right)^{a-1}}$

Hint:

Use chain rule and   $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given:

\begin{aligned} &x=\left(t+\frac{1}{t}\right)^{a} \\ &y=(a)^{t+\frac{1}{t}} \end{aligned}

Solution:

$x=\left(t+\frac{1}{t}\right)^{a} \\$

\begin{aligned} & &\frac{d x}{d t}=\frac{d\left(t+\frac{1}{t}\right)}{d t} \end{aligned}

$=\frac{d\left(t+\frac{1}{t}\right)^{a}}{d\left(t+\frac{1}{t}\right)} \times \frac{d\left(t+\frac{1}{t}\right)}{d t}$                      [Using chain rule]

$=a\left(t+\frac{1}{t}\right)^{a-1} \times\left(\frac{d t}{d t}+\frac{d\left(\frac{1}{t}\right)}{d t}\right)$                  $\left[\because \frac{d x^{n}}{d x}=n x^{n-1}\right]$

$=a\left(t+\frac{1}{t}\right)^{a-1} \times\left(1+\left(\frac{-1}{t^{2}}\right)\right) \\$

\begin{aligned} & &\frac{d x}{d t}=a\left(t+\frac{1}{t}\right)^{a-1} \times\left(1-\frac{1}{t^{2}}\right) \end{aligned}                                                                                                       (1)

$y=a^{t+\frac{1}{t}} \\$

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

\begin{aligned} & &=\frac{d\left(a^{t+\frac{1}{t}}\right)}{d\left(t+\frac{1}{t}\right)} \times \frac{d\left(t+\frac{1}{t}\right)}{d t} \end{aligned}

$=\left(a^{t+\frac{1}{t}}\right) \log a \times\left(\frac{d t}{d t}+\frac{d\left(\frac{1}{t}\right)}{d t}\right)$                          $\quad\left[\because \frac{d\left(a^{x}\right)}{d x}=a^{x} \log a\right]$

$\frac{d y}{d t}=(a)^{t+\frac{1}{t}} \log a \times\left[1-\frac{1}{t^{2}}\right]$                                                                                                                                        (2)

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

Put the values of  $\frac{d x}{d t} \text { and } \frac{d y}{d t}$  from equation (1) and (2) respectively

$\frac{d y}{d x}=\frac{(a)^{r+\frac{1}{1}} \times \log a \times\left(1-\frac{1}{t^{2}}\right)}{a\left(t+\frac{1}{t}\right)^{a-1} \times\left(1-\frac{1}{t^{2}}\right)} \\$

\begin{aligned} & &\frac{d y}{d x}=\frac{(a)^{t+\frac{1}{t}} \log a}{a\left(t+\frac{1}{t}\right)^{a-1}} \end{aligned}