Find dy - dx in the following questions-x y-y2-tan x-y

Given, xy+y^2=tan x+y Differentiating both sides w.r.t. x, we get d/dx(xy+y^2)=d/dx(tan x+y) d/dx(xy)+d/dx(y^2)=sec^2x+dy/dx ⇒ xdy/dx+2 y dy/dx=sec^2x+dy/dx ...

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Byju's Answer

Standard XII

Mathematics

Formation of a Differential Equation from a General Solution

Find dy / dx ...

Question

Find $\frac{d y}{d x}$ in the following questions:

$x y + y^{2} = t a n x + y$

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Solution

Given, $x y + y^{2} = t a n x + y$

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

$\frac{d}{d x} (x y + y^{2}) = \frac{d}{d x} (t a n x + y)$

$\frac{d}{d x} (x y) + \frac{d}{d x} (y^{2}) = s e c^{2} x + \frac{d y}{d x}$

$\Rightarrow x \frac{d y}{d x} + 2 y \frac{d y}{d x} = s e c^{2} x + \frac{d y}{d x}$

$(U s i n g p r o d u c t r u l e \frac{d}{d x} (u . v) = u \frac{d}{d x} v + v \frac{d}{d x} u)$

$\Rightarrow x \frac{d y}{d x} + 2 y \frac{d y}{d x} - \frac{d y}{d x} = s e c^{2} x - y \Rightarrow (x + 2 y - 1) \frac{d y}{d x} = s e c^{2} x - y$

$\Rightarrow \frac{d y}{d x} = \frac{s e c^{2} x - y}{x + 2 y - 1}$

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Find dydxin the following questions: xy+y2=tan x+y

Given, xy+y2=tan x+y Differentiating both sides w.r.t. x, we get ddx(xy+y2)=ddx(tan x+y) ddx(xy)+ddx(y2)=sec2x+dydx ⇒ xdydx+2 y dydx=sec2x+dydx (Using product ruleddx(u.v)=uddxv+vddxu) ⇒ xdydx+2ydydx−dydx=sec2x−y⇒(x+2y−1)dydx=sec2x−y ⇒ dydx=sec2x−yx+2y−1

Find $\frac{d y}{d x}$ in the following questions:

$x y + y^{2} = t a n x + y$