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8

5quad The variables 
x and 
y are such that 
y=0 when 
x=0 and

(x+1)y+(x+y+1)^(3)=1.
(a) Show that 
(dy)/((d)x)=-(3)/(4) when 
x=0.

88\newline5 5 \quad The variables x x and y y are such that y=0 y=0 when x=0 x=0 and\newline(x+1)y+(x+y+1)3=1. (x+1) y+(x+y+1)^{3}=1 . \newline(a) Show that dy dx=34 \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{3}{4} when x=0 x=0 .

Full solution

Q. 88\newline5 5 \quad The variables x x and y y are such that y=0 y=0 when x=0 x=0 and\newline(x+1)y+(x+y+1)3=1. (x+1) y+(x+y+1)^{3}=1 . \newline(a) Show that dy dx=34 \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{3}{4} when x=0 x=0 .
  1. Given Equation and Condition: Given the equation (x+1)y+(x+y+1)3=1(x+1)y + (x+y+1)^3 = 1 and the condition y=0y=0 when x=0x=0.
  2. Implicit Differentiation: Differentiate both sides with respect to xx using implicit differentiation: ddx[(x+1)y]+ddx[(x+y+1)3]=ddx[1]\frac{d}{dx}[(x+1)y] + \frac{d}{dx}[(x+y+1)^3] = \frac{d}{dx}[1]. For the first term, apply the product rule: (x+1)dydx+y(x+1)\frac{dy}{dx} + y. For the second term, apply the chain rule: 3(x+y+1)2(1+dydx)3(x+y+1)^2 * (1 + \frac{dy}{dx}). Set the derivative of the constant (11) to 00.
  3. Substitute and Simplify: Substitute y=0y=0 and x=0x=0 into the differentiated equation: (0+1)dydx+0+3(0+0+1)2(1+dydx)=0(0+1)\frac{dy}{dx} + 0 + 3(0+0+1)^2 \cdot (1 + \frac{dy}{dx}) = 0. Simplify to get: dydx+3(1)(1+dydx)=0\frac{dy}{dx} + 3(1) \cdot (1 + \frac{dy}{dx}) = 0. This simplifies to dydx+3+3dydx=0\frac{dy}{dx} + 3 + 3\frac{dy}{dx} = 0.
  4. Combine and Solve: Combine like terms and solve for dydx\frac{dy}{dx}: 4dydx+3=04\frac{dy}{dx} + 3 = 0. Subtract 33 from both sides: 4dydx=34\frac{dy}{dx} = -3. Divide by 44: dydx=34\frac{dy}{dx} = -\frac{3}{4}.

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