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Find the derivative dy/dx for the following equations: 1) x2 + y2 = x + 4 + 8. 2) x3 + y3 = x3 y3 - 1.

Mathematics

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Here are the differentiations with respect to xx:

  1. For the equation x2+y2=x+4+8x^2 + y^2 = x + 4 + 8:

  1. Simplify the equation.

    x2+y2=x+12x^2 + y^2 = x + 12

  2. Differentiate both sides with respect to xx.

    Remember to use the chain rule for terms involving yy.

    ddx(x2)+ddx(y2)=ddx(x)+ddx(12)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(x) + \frac{d}{dx}(12) 2x+2ydydx=1+02x + 2y \frac{dy}{dx} = 1 + 0

  3. Isolate dydx\frac{dy}{dx}.

    2ydydx=1−2x2y \frac{dy}{dx} = 1 - 2x dydx=1−2x2y\frac{dy}{dx} = \frac{1 - 2x}{2y}

    The derivative is:

    dydx=1−2x2y\boxed{\frac{dy}{dx} = \frac{1 - 2x}{2y}}
    1. For the equation x3+y3=x3y3−1x^3 + y^3 = x^3 y^3 - 1:

  4. Differentiate both sides with respect to xx.

    Use the chain rule for yy terms and the product rule for x3y3x^3 y^3.

    ddx(x3)+ddx(y3)=ddx(x3y3)−ddx(1)\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(x^3 y^3) - \frac{d}{dx}(1) 3x2+3y2dydx=(3x2y3+x3⋅3y2dydx)−03x^2 + 3y^2 \frac{dy}{dx} = (3x^2 y^3 + x^3 \cdot 3y^2 \frac{dy}{dx}) - 0 3x2+3y2dydx=3x2y3+3x3y2dydx3x^2 + 3y^2 \frac{dy}{dx} = 3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx}

  5. Group terms containing dydx\frac{dy}{dx} on one side and other terms on the other side.

    3y2dydx−3x3y2dydx=3x2y3−3x23y^2 \frac{dy}{dx} - 3x^3 y^2 \frac{dy}{dx} = 3x^2 y^3 - 3x^2

  6. Factor out dydx\frac{dy}{dx} and solve for it.

    dydx(3y2−3x3y2)=3x2y3−3x2\frac{dy}{dx} (3y^2 - 3x^3 y^2) = 3x^2 y^3 - 3x^2 dydx=3x2y3−3x23y2−3x3y2\frac{dy}{dx} = \frac{3x^2 y^3 - 3x^2}{3y^2 - 3x^3 y^2}

  7. Simplify the expression by factoring out common terms.

    dydx=3x2(y3−1)3y2(1−x3)\frac{dy}{dx} = \frac{3x^2 (y^3 - 1)}{3y^2 (1 - x^3)} dydx=x2(y3−1)y2(1−x3)\frac{dy}{dx} = \frac{x^2 (y^3 - 1)}{y^2 (1 - x^3)}

    The derivative is:

    dydx=x2(y3−1)y2(1−x3)\boxed{\frac{dy}{dx} = \frac{x^2 (y^3 - 1)}{y^2 (1 - x^3)}}
    1. For the equation xy=yxx^y = y^x:

  8. Take the natural logarithm of both sides to bring down the exponents.

    ln⁡(xy)=ln⁡(yx)\ln(x^y) = \ln(y^x) yln⁡x=xln⁡yy \ln x = x \ln y

  9. Differentiate both sides with respect to xx using the product rule.

    ddx(yln⁡x)=ddx(xln⁡y)\frac{d}{dx}(y \ln x) = \frac{d}{dx}(x \ln y) (dydx⋅ln⁡x+y⋅1x)=(1⋅ln⁡y+x⋅1ydydx)\left(\frac{dy}{dx} \cdot \ln x + y \cdot \frac{1}{x}\right) = \left(1 \cdot \ln y + x \cdot \frac{1}{y} \frac{dy}{dx}\right) ln⁡xdydx+yx=ln⁡y+xydydx\ln x \frac{dy}{dx} + \frac{y}{x} = \ln y + \frac{x}{y} \frac{dy}{dx}

  10. Group terms containing dydx\frac{dy}{dx} on one side and other terms on the other side.

    ln⁡xdydx−xydydx=ln⁡y−yx\ln x \frac{dy}{dx} - \frac{x}{y} \frac{dy}{dx} = \ln y - \frac{y}{x}

  11. Factor out dydx\frac{dy}{dx} and solve for it.

    dydx(ln⁡x−xy)=ln⁡y−yx\frac{dy}{dx} \left(\ln x - \frac{x}{y}\right) = \ln y - \frac{y}{x} dydx(yln⁡x−xy)=xln⁡y−yx\frac{dy}{dx} \left(\frac{y \ln x - x}{y}\right) = \frac{x \ln y - y}{x} dydx=xln⁡y−yxyln⁡x−xy\frac{dy}{dx} = \frac{\frac{x \ln y - y}{x}}{\frac{y \ln x - x}{y}} dydx=xln⁡y−yx⋅yyln⁡x−x\frac{dy}{dx} = \frac{x \ln y - y}{x} \cdot \frac{y}{y \ln x - x} dydx=y(xln⁡y−y)x(yln⁡x−x)\frac{dy}{dx} = \frac{y(x \ln y - y)}{x(y \ln x - x)}

    The derivative is:

    dydx=y(xln⁡y−y)x(yln⁡x−x)\boxed{\frac{dy}{dx} = \frac{y(x \ln y - y)}{x(y \ln x - x)}}
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