Question
If yx = ey–x, prove that $\frac{\text{dy}}{\text{dx}} = \frac{(1 + \log\text{y})^{2}}{\log\text{y}}.$
Taking logarithm both sides we get
log
yx = log ey-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 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}.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Machine | Area occupied | Labour force | Daily output ( in units) |
| A | 1000m2 | 12 men | 60 |
| B | 1200m2 | 8 men | 40 |
He has
maximum area of 9000 $\text{m}^{2}$ available, and 72 skilled labourers who can operate both the machines. How many machines of each type should he buy to maximise the daily output?$(\text{x}^{2} - 1 ) \frac{\text{dy}}{\text{dx}} + 2 \text{xy} = \frac{2}{\text{x}^{2} - 1 }.$