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Name: provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 10
Uploaded: Wed, 2021-08-04 21:29
Description: provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 10

provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 10

Answers (1)

Answer: $\frac{y}{x}\left[\frac{x e^{(x-y)}-1}{y e^{(x-y)}-1}\right]$

Hint:

Use chain rule and quotient rule

Given:

$e^{x-y}=\log \left(\frac{x}{y}\right)$

Solution:

Differentiate the given equation w.r.t x

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

$\frac{d\left(e^{x-y}\right)}{d(x-y)} \times \frac{d(x-y)}{d x}=\frac{d\left(\log \left(\frac{x}{y}\right)\right)}{d\left(\frac{x}{y}\right)} \times \frac{d\left(\frac{x}{y}\right)}{d x}$

$\left(e^{x-y}\right) \times\left[\frac{d x}{d x}-\frac{d y}{d x}\right]=\frac{1}{\left(\frac{x}{y}\right)} \times\left[\frac{y \cdot \frac{d x}{d x}-x \cdot \frac{d y}{d x}}{y^{2}}\right]$ $\left[\because \frac{d\left(e^{x}\right)}{d x}=e^{x}\right]$