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

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

Answers (1)

Answer:

            \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}

Posted by

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