# How do you implicitly differentiate 2=e^(x-y)/cosy ?

Jul 6, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 - \tan y} .$

#### Explanation:

Writing the given eqn. as, $2 \cos y = {e}^{x - y}$
$\therefore \ln \left(2 \cos y\right) = \ln \left({e}^{x - y}\right) .$ Using the Rules of Log., we have,
$\therefore \ln 2 + \ln \cos y = \left(x - y\right) \ln e ,$ i.e., $\ln 2 + \ln \cos y = x - y$

Now we diff. both sides w.r.t. $x ,$

$\frac{d}{\mathrm{dx}} \ln 2 + \frac{d}{\mathrm{dx}} \left(\ln \cos y\right) = \frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} \left(y\right)$

:. $0 + \frac{d}{\mathrm{dy}} \left(\ln \cos y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$..................... [Chain Rule]
$\therefore \frac{1}{\cos} y \cdot \frac{d}{\mathrm{dy}} \left(\cos y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore \left(\frac{1}{\cos} y\right) \left(- \sin y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore - \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore \left(1 - \tan y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 - \tan y} .$

Alternatively, $2 \cos y = {e}^{x - y} \ldots \left(1\right) \Rightarrow 2 \cos y \cdot {e}^{y} = {e}^{x}$
Diif.ing both sides, w.r.t. $y$, we have,
$2 \left\{{e}^{y} \cdot \cos y - {e}^{y} \cdot \sin y\right\} = \frac{d}{\mathrm{dy}} {e}^{x} = \frac{d}{\mathrm{dx}} {e}^{x} \cdot \frac{\mathrm{dx}}{\mathrm{dy}} = {e}^{x} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}$
$\therefore 2 {e}^{y} \left(\cos y - \sin y\right) = {e}^{x} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}$
$\therefore 2 \left(\cos y - \sin y\right) = {e}^{x - y} \frac{\mathrm{dx}}{\mathrm{dy}} = 2 \cos y \frac{\mathrm{dx}}{\mathrm{dy}} ,$.........[.by $\left(1\right)$]
Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} = \cos \frac{y}{\cos y - \sin y}$, or, $= \frac{1}{1 - \tan y} ,$
as obtained earlier!

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