What is the slope of the tangent line of $\frac{{(x + 2 y)}^{2}}{e^{x - y^{2}}} = C$ $(x+2y)^2/(e^(x-y^2)) =C$ , where C is an arbitrary constant, at $(1, 1)$ $(1,1)$ ?

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Answer 1 · 2018-05-03T16:33:41

$m = \frac{1}{10}$ $m = 1/10$

Given that the curve contains the point $(1, 1)$ $(1,1)$ , we find $C = 9$ $C = 9$

The equation is equivalent to

$x^{2} + 4 x y + 4 y^{2} = 9 e^{x - y^{2}}$ $x^2+4xy+4y^2 = 9e^(x-y^2)$ .

Differentiate implicitly to get

$2 x + 4 y + 4 x \frac{d y}{d x} + 8 y \frac{d y}{d x} = 9 e^{x - y^{2}} (1 - 2 y \frac{d y}{d x})$ $2x+4y+4xdy/dx+8ydy/dx=9e^(x-y^2)(1-2ydy/dx)$

At $(1, 1)$ $(1,1)$ we have

$2 + 4 + 4 \frac{d y}{d x} + 8 \frac{d y}{d x} = 9 - 18 \frac{d y}{d x}$ $2+4+4dy/dx+8dy/dx=9-18dy/dx$

so, $\frac{d y}{d x} = \frac{1}{10}$ $dy/dx = 1/10$

What is the slope of the tangent line of (x+2y)^2/(e^(x-y^2)) =C (x+2y)2ex−y2=C (x+2y)^2/(e^(x-y^2)) =C , where C is an arbitrary constant, at (1,1)(1,1)(1,1)?