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The equation of a curve is y=2-x-(2x+3)/(x-3). i) Find (dy)/((d)x) and (d^(2)y)/((d)x^(2)).

The equation of a curve is y=2−x−2x+3x−3 y=2-x-\frac{2 x+3}{x-3} .\newlinei) Find dy dx \frac{\mathrm{d} y}{\mathrm{~d} x} and d2y dx2 \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} .

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Q. The equation of a curve is y=2−x−2x+3x−3 y=2-x-\frac{2 x+3}{x-3} .\newlinei) Find dy dx \frac{\mathrm{d} y}{\mathrm{~d} x} and d2y dx2 \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} .
  1. Find First Derivative: Now, let's find the first derivative (dydx)(\frac{dy}{dx}).(dydx)=ddx(−x)−ddx(2xx−3)+ddx(3x−3)(\frac{dy}{dx}) = \frac{d}{dx}(-x) - \frac{d}{dx}(\frac{2x}{x - 3}) + \frac{d}{dx}(\frac{3}{x - 3})(dydx)=−1−(2(x−3)−2x⋅1)(x−3)2+(3⋅1)(x−3)2(\frac{dy}{dx}) = -1 - \frac{(2(x - 3) - 2x \cdot 1)}{(x - 3)^2} + \frac{(3 \cdot 1)}{(x - 3)^2}(dydx)=−1−(2x−6−2x)(x−3)2+3(x−3)2(\frac{dy}{dx}) = -1 - \frac{(2x - 6 - 2x)}{(x - 3)^2} + \frac{3}{(x - 3)^2}(dydx)=−1−(−6)(x−3)2+3(x−3)2(\frac{dy}{dx}) = -1 - \frac{(-6)}{(x - 3)^2} + \frac{3}{(x - 3)^2}(dydx)=−1+9(x−3)2(\frac{dy}{dx}) = -1 + \frac{9}{(x - 3)^2}
  2. Calculate Second Derivative: Next, let's find the second derivative (d2ydx2)(\frac{d^2y}{dx^2}).
    (d2ydx2)=ddx(−1+9(x−3)2)(\frac{d^2y}{dx^2}) = \frac{d}{dx}(-1 + \frac{9}{(x - 3)^2})
    (d2ydx2)=ddx(−1)+ddx(9(x−3)2)(\frac{d^2y}{dx^2}) = \frac{d}{dx}(-1) + \frac{d}{dx}(\frac{9}{(x - 3)^2})
    (d2ydx2)=0+9⋅ddx(1(x−3)2)(\frac{d^2y}{dx^2}) = 0 + 9 \cdot \frac{d}{dx}(\frac{1}{(x - 3)^2})
    (d2ydx2)=9⋅(−2)⋅(1(x−3)3)⋅ddx(x−3)(\frac{d^2y}{dx^2}) = 9 \cdot (-2) \cdot (\frac{1}{(x - 3)^3}) \cdot \frac{d}{dx}(x - 3)
    (d2ydx2)=−18(x−3)3(\frac{d^2y}{dx^2}) = -\frac{18}{(x - 3)^3}

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