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Implicite function

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Answer

dydx=1−4x−yexy2y+xexy\frac{dy}{dx} = \frac{1 - 4x - y e^{xy}}{2y + x e^{xy}}

Step-by-step solution

6 steps

To find dydx\frac{dy}{dx} using implicit differentiation, we differentiate every term in the equation with respect to xx, remembering to apply the chain rule for terms involving yy.

Given the equation:

2x2+y2+exy=x2x^2 + y^2 + e^{xy} = x

  1. Differentiate each term with respect to xx.

    ddx(2x2)+ddx(y2)+ddx(exy)=ddx(x)\frac{d}{dx}(2x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(e^{xy}) = \frac{d}{dx}(x)

  2. Apply the differentiation rules.

    • For 2x22x^2: ddx(2x2)=4x\frac{d}{dx}(2x^2) = 4x.
    • For y2y^2: Using the chain rule, ddx(y2)=2ydydx\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}.
    • For exye^{xy}: Using the chain rule and product rule for the exponent xyxy: ddx(exy)=exy⋅ddx(xy)\frac{d}{dx}(e^{xy}) = e^{xy} \cdot \frac{d}{dx}(xy) ddx(xy)=(1⋅y)+(x⋅dydx)=y+xdydx\frac{d}{dx}(xy) = (1 \cdot y) + (x \cdot \frac{dy}{dx}) = y + x \frac{dy}{dx} So, ddx(exy)=exy(y+xdydx)=yexy+xexydydx\frac{d}{dx}(e^{xy}) = e^{xy}(y + x \frac{dy}{dx}) = y e^{xy} + x e^{xy} \frac{dy}{dx}.
    • For xx: ddx(x)=1\frac{d}{dx}(x) = 1.

  3. Substitute these derivatives back into the equation.

    4x+2ydydx+yexy+xexydydx=14x + 2y \frac{dy}{dx} + y e^{xy} + x e^{xy} \frac{dy}{dx} = 1

  4. Group all terms containing dydx\frac{dy}{dx} on one side of the equation and move all other terms to the other side.

    2ydydx+xexydydx=1−4x−yexy2y \frac{dy}{dx} + x e^{xy} \frac{dy}{dx} = 1 - 4x - y e^{xy}

  5. Factor out dydx\frac{dy}{dx} from the terms on the left side.

    dydx(2y+xexy)=1−4x−yexy\frac{dy}{dx} (2y + x e^{xy}) = 1 - 4x - y e^{xy}

  6. Solve for dydx\frac{dy}{dx} by dividing both sides by (2y+xexy)(2y + x e^{xy}).

    dydx=1−4x−yexy2y+xexy\frac{dy}{dx} = \boxed{\frac{1 - 4x - y e^{xy}}{2y + x e^{xy}}}
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