Find the second derivative of y = (x2+x+1)2
Solution;
Given;
y=(x2+x+1)2y=(x^2+x+1)^2y=(x2+x+1)2
dydx=2(x2+x+1)ddx(x2+x+1)\frac{dy}{dx}=2(x^2+x+1)\frac{d}{dx}(x^2+x+1)dxdy=2(x2+x+1)dxd(x2+x+1)
y′=2(x2+x+1)(2x+1)y'=2(x^2+x+1)(2x+1)y′=2(x2+x+1)(2x+1)
Apply product rule;
y′′=2(x2+x+1)ddx(2x+1)+2(2x+1)ddx(x2+x+1)y''=2(x^2+x+1)\frac{d}{dx}(2x+1)+2(2x+1)\frac{d}{dx}(x^2+x+1)y′′=2(x2+x+1)dxd(2x+1)+2(2x+1)dxd(x2+x+1)
y′′=4(x2+x+1)+2(2x+1)(2x+1)y''=4(x^2+x+1)+2(2x+1)(2x+1)y′′=4(x2+x+1)+2(2x+1)(2x+1)
y′′=4(x2+x+1)+2(2x+1)2y''=4(x^2+x+1)+2(2x+1)^2y′′=4(x2+x+1)+2(2x+1)2
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