Q. 85 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 dxdy=−43 when x=0.
Given Equation and Condition: Given the equation (x+1)y+(x+y+1)3=1 and the condition y=0 when x=0.
Implicit Differentiation: Differentiate both sides with respect to x using implicit differentiation: dxd[(x+1)y]+dxd[(x+y+1)3]=dxd[1]. For the first term, apply the product rule: (x+1)dxdy+y. For the second term, apply the chain rule: 3(x+y+1)2∗(1+dxdy). Set the derivative of the constant (1) to 0.
Substitute and Simplify: Substitute y=0 and x=0 into the differentiated equation: (0+1)dxdy+0+3(0+0+1)2⋅(1+dxdy)=0. Simplify to get: dxdy+3(1)⋅(1+dxdy)=0. This simplifies to dxdy+3+3dxdy=0.
Combine and Solve: Combine like terms and solve for dxdy: 4dxdy+3=0. Subtract 3 from both sides: 4dxdy=−3. Divide by 4: dxdy=−43.
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