Question

find dy/dx

y=tan xy???????​

Answer 1

Step-by-step explanation:

We have,

$y = \tan(xy)$

Differentiating both sides w. r. t x, we get,

$\frac{dy}{dx} = \sec^{2} (xy). \frac{d}{dx} (xy)$

$\implies \frac{dy}{dx} = \sec^{2} (xy). \bigg \{y \frac{d}{dx} (x) + x \frac{d}{dx} (y)\bigg \}$

$\implies \frac{dy}{dx} = \sec^{2} (xy). \bigg \{y+ x \frac{dy}{dx}\bigg \}$

$\implies \frac{dy}{dx} = y\sec^{2} (xy) + x \sec^{2} (xy) \frac{dy}{dx}$

$\implies \frac{dy}{dx} - x \sec^{2} (xy) \frac{dy}{dx} = y\sec^{2} (xy)$

$\implies \bigg \{1- x \sec^{2} (xy) \bigg \} \frac{dy}{dx} = y\sec^{2} (xy)$

$\implies \frac{dy}{dx} = \frac{y\sec^{2} (xy) }{ 1- x \sec^{2} (xy) }$