How do you implicitly differentiate $2 = y tan x y - x y$ $2=ytanxy-xy$ ?

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How do you implicitly differentiate $2 = y tan x y - x y$ $2=ytanxy-xy$ ?

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Answer 1 · 2017-08-06T09:11:06

$\frac{d y}{d x} = \frac{y - y^{2} {sec}^{2} (x y)}{x y {sec}^{2} (x y) + tan (x y) - x}$ $dy/dx=(y-y^2sec^2(xy))/(xysec^2(xy)+tan(xy)-x)$

Explanation:

$differentiate implicitly with respect to x$ $"differentiate "color(blue)"implicitly with respect to x"$

$differentiate y tan (x y) using the product rule$ $"differentiate "ytan(xy)" using the "color(blue)"product rule"$

$\frac{d}{d x} (y tan (x y))$ $d/dx(ytan(xy))$

$= y {sec}^{2} (x y) . (x \frac{d y}{d x} + y) + tan (x y) . \frac{d y}{d x}$ $=ysec^2(xy).(xdy/dx+y)+tan(xy).dy/dx$

$= x y {sec}^{2} (x y) \frac{d y}{d x} + y^{2} {sec}^{2} (x y) + tan (x y) \frac{d y}{d x}$ $=xysec^2(xy)dy/dx+y^2sec^2(xy)+tan(xy)dy/dx$

$differentiate - x y using the product rule$ $"differentiate "-xy" using the "color(blue)"product rule"$

$\frac{d}{d x} (- x y) = - x \frac{d y}{d x} - y$ $d/dx(-xy)=-xdy/dx-y$

$putting the right side all together$ $"putting the right side all together"$

$x y {sec}^{2} (x y) \frac{d y}{d x} + y^{2} {sec}^{2} (x y) + tan (x y) \frac{d y}{d x} - x \frac{d y}{d x} - y$ $xysec^2(xy)dy/dx+y^2sec^2(xy)+tan(xy)dy/dx-xdy/dx-y$

$= \frac{d y}{d x} (x y {sec}^{2} (x y) + tan (x y) - x) + y^{2} {sec}^{2} (x y) - y = 0$ $=dy/dx(xysec^2(xy)+tan(xy)-x)+y^2sec^2(xy)-y=0$

$\Rightarrow \frac{d y}{d x} = \frac{y - y^{2} {sec}^{2} (x y)}{x y {sec}^{2} (x y) + tan (x y) - x}$ $rArrdy/dx=(y-y^2sec^2(xy))/(xysec^2(xy)+tan(xy)-x)$

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How do you implicitly differentiate $2 = y tan x y - x y$ $2=ytanxy-xy$ ?

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