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Question Number 123667 by sahnaz last updated on 27/Nov/20

$$\mathrm{y}={(\mathrm{x}+1)}^3{(\mathrm{x}-1)}^2{\mathrm{d}}^2\mathrm{y}=?$$

Answered by mathmax by abdo last updated on 27/Nov/20

$$\mathrm{y}\left(\mathrm{x}\right)={(\mathrm{x}+1)}^3{(\mathrm{x}-1)}^2\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=(\mathrm{x}+1){({\mathrm{x}}^2-1)}^2\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}={({\mathrm{x}}^2-1)}^2+2\left(2\mathrm{x}\right)({\mathrm{x}}^2-1)(\mathrm{x}+1)={({\mathrm{x}}^2-1)}^2+4\mathrm{x}(\mathrm{x}+1)({\mathrm{x}}^2-1)$$$$={({\mathrm{x}}^2-1)}^2+4({\mathrm{x}}^2+\mathrm{x})({\mathrm{x}}^2-1)\Rightarrow$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2\left(2\mathrm{x}\right)({\mathrm{x}}^2-1)+4\{(2\mathrm{x}+1)({\mathrm{x}}^2-1)+2\mathrm{x}({\mathrm{x}}^2+\mathrm{x})\}$$$$=4\mathrm{x}({\mathrm{x}}^2-1)+4\{{2\mathrm{x}}^3-2\mathrm{x}+{\mathrm{x}}^2-1+{2\mathrm{x}}^3+{2\mathrm{x}}^2\}$$$$={4\mathrm{x}}^3-4\mathrm{x}+4\{{4\mathrm{x}}^3+{3\mathrm{x}}^2-2\mathrm{x}-1\}$$$$={4\mathrm{x}}^3-4\mathrm{x}+{16\mathrm{x}}^3+{12\mathrm{x}}^2+8\mathrm{x}-4$$$$={20\mathrm{x}}^3+{12\mathrm{x}}^2+4\mathrm{x}-4$$

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