Heidi Tate

2023-03-11

How to find dy/dx if $x+\mathrm{tan}\left(xy\right)=0$?

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Beginner2023-03-12Added 3 answers

Step 1

You must remember that y is a function of x and derive it accordingly, as $\frac{dy}{dx}$ getting:

$x+\mathrm{tan}(xy)=0\phantom{\rule{0ex}{0ex}}1+\frac{1}{{\mathrm{cos}}^{2}(xy)}\cdot [y+x\frac{dy}{dx}]=0\phantom{\rule{0ex}{0ex}}1+\frac{y}{{\mathrm{cos}}^{2}(xy)}+\frac{x}{{\mathrm{cos}}^{2}(xy)}\frac{dy}{dx}=0$

Step 2

$\frac{dy}{dx}=(-1-\frac{y}{{\mathrm{cos}}^{2}(xy)})\cdot \frac{{\mathrm{cos}}^{2}(xy)}{x}=\frac{1}{x}({\mathrm{cos}}^{2}(xy)-y)$

You must remember that y is a function of x and derive it accordingly, as $\frac{dy}{dx}$ getting:

$x+\mathrm{tan}(xy)=0\phantom{\rule{0ex}{0ex}}1+\frac{1}{{\mathrm{cos}}^{2}(xy)}\cdot [y+x\frac{dy}{dx}]=0\phantom{\rule{0ex}{0ex}}1+\frac{y}{{\mathrm{cos}}^{2}(xy)}+\frac{x}{{\mathrm{cos}}^{2}(xy)}\frac{dy}{dx}=0$

Step 2

$\frac{dy}{dx}=(-1-\frac{y}{{\mathrm{cos}}^{2}(xy)})\cdot \frac{{\mathrm{cos}}^{2}(xy)}{x}=\frac{1}{x}({\mathrm{cos}}^{2}(xy)-y)$