If y^x = e^ y–x , prove that dy dx = (1 + )^ 2 . - Vidyadip

Question

If $y^x = e^{y–x},$ prove that $\frac{\text{dy}}{\text{dx}} = \frac{(1 + \log\text{y})^{2}}{\log\text{y}}.$

✓

Answer

Given $y^x = e^{y-x}$
Taking logarithm both sides we get
$\log y^x = \log e^{y-x}$
$\Rightarrow\text{x}.\log\text{y} = (\text{y} - \text{x}).\log e$
$\Rightarrow\text{x}.\log\text{y} = (\text{y} - \text{x})$
$\Rightarrow\text{x}(1 + \log\text{y}) = \text{y}$
$\Rightarrow\text{x} = \frac{\text{y}}{1 + \log\text{y}}$
Differentiating both sides $\text{w.r.t.y.}$ We get
$\frac{\text{dx}}{\text{dy}} = \frac{(1 + \log\text{y}).1 - \text{y}.\bigg(0 + \frac{1}{\text{y}}\bigg)}{(1 + \log\text{y})^{2}}$
$ = \frac{1 + \log\text{y} - 1 }{(1 + \log\text{y})^{2}} = \frac{\log\text{y}}{(1 + \log\text{y})^{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{(1 + \log\text{y})^{2}}{\log\text{y}}$
$ \begin{bmatrix} \text{Note}:(i) \log_{e} \text{mn} = \log_{e}\text{m} + \log_{e}\text{n} \\ (ii)\log_{e}\frac{\text{m}}{\text{n}} = \log_{e}\text{m} - \log_{e}\text{n}\\ (iii)\log_{e } \text{ m}^{n} = \text{n}\log_{e}\text{m} \end{bmatrix}.$

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