Let u=(x-1) -> (du)/dx=1u=(x−1)→dudx=1
Let v=(x^2+2)^3->(du)/dx=3(x^2+2)^2(2x) =6x(x^2+2)^2v=(x2+2)3→dudx=3(x2+2)2(2x)=6x(x2+2)2
Set y=(x-1)(x^2+2)^3 ->uvy=(x−1)(x2+2)3→uv
Using dy/dx=v(du)/dx+u(dv)/dxdydx=vdudx+udvdx
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dy/dx" "=" "[(x^2+2)^3xx1] + (x-1)[6x(x^2+2)^2] dydx = [(x2+2)3×1]+(x−1)[6x(x2+2)2]......Eqn(1)
=>dy/dx" "=" "(x^2+2)^2[(x^2+2)+6x(x-1)]⇒dydx = (x2+2)2[(x2+2)+6x(x−1)]
=>dy/dx" "=" "(x^2+2)^2(7x^2-6x+2)⇒dydx = (x2+2)2(7x2−6x+2)
=>dy/dx" "=" "(x^4+4x^2+4)(7x^2-6x+2)⇒dydx = (x4+4x2+4)(7x2−6x+2)
dy/dx=color(blue)(7x^6 -6x^5+2x^4)color(green)(+28x^4-24x^3+8x^2)color(purple)(+28x^2-24x+8)dydx=7x6−6x5+2x4+28x4−24x3+8x2+28x2−24x+8
dy/dx=7x^6-6x^5+30x^4-24x^3+ 36x^2-24x+8dydx=7x6−6x5+30x4−24x3+36x2−24x+8
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