What is the implicit derivative of $1 = tan y x - y^{2}$ $1=tanyx-y^2$ ?

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What is the implicit derivative of $1 = tan y x - y^{2}$ $1=tanyx-y^2$ ?

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Answer 1 · 2017-09-28T13:30:07

$\frac{d y}{d x} = \frac{y {sec}^{2} (y x)}{2 y - x {sec}^{2} (y x)}$ $dy/dx=(ysec^2(yx))/(2y-xsec^2(yx))$

Explanation:

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

$differentiate tan (y x) using chain/product rules$ $"differentiate "tan(yx)" using "color(blue)"chain/product rules"$

$0 = {sec}^{2} (y x) \times \frac{d}{d x} (y x) - 2 y \frac{d y}{d x}$ $0=sec^2(yx)xxd/dx(yx)-2ydy/dx$

$\Rightarrow 2 y \frac{d y}{d x} = {sec}^{2} (y x) (y + x \frac{d y}{d x})$ $rArr2ydy/dx=sec^2(yx)(y+xdy/dx)$

$\Rightarrow 2 y \frac{d y}{d x} = y {sec}^{2} (y x) + x {sec}^{2} (y x) \frac{d y}{d x}$ $color(white)(rArr2ydy/dx)=ysec^2(yx)+xsec^2(yx)dy/dx$

$\Rightarrow 2 y \frac{d y}{d x} - x {sec}^{2} (y x) \frac{d y}{d x} = y {sec}^{2} (y x)$ $rArr2ydy/dx-xsec^2(yx)dy/dx=ysec^2(yx)$

$\Rightarrow \frac{d y}{d x} (2 y - x {sec}^{2} (y x)) = y {sec}^{2} (y x)$ $rArrdy/dx(2y-xsec^2(yx))=ysec^2(yx)$

$\Rightarrow \frac{d y}{d x} = \frac{y {sec}^{2} (y x)}{2 y - x {sec}^{2} (y x)}$ $rArrdy/dx=(ysec^2(yx))/(2y-xsec^2(yx))$

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What is the implicit derivative of $1 = tan y x - y^{2}$ $1=tanyx-y^2$ ?

1 Answer

Explanation:

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