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Question Number 144525 by imjagoll last updated on 26/Jun/21

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

Answered by EDWIN88 last updated on 26/Jun/21

$$\mathrm{let}\mathrm{q}={\mathrm{x}}^2-3\mathrm{x}+1\to \mathrm{y}=\cosh \mathrm{q}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dq}}.\frac{\mathrm{dq}}{\mathrm{dx}}=\left(\sinh \mathrm{q}\right)(2\mathrm{x}-3)$$$$=(2\mathrm{x}-3)\sinh ({\mathrm{x}}^2-3\mathrm{x}+1)$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sinh \mathrm{q}\frac{\mathrm{dq}}{\mathrm{dx}}\right)$$$$=\sinh \mathrm{q}\frac{{\mathrm{d}}^2\mathrm{q}}{{\mathrm{dx}}^2}+\cosh \mathrm{q}{\left(\frac{\mathrm{dq}}{\mathrm{dx}}\right)}^2$$$$=2\sinh \mathrm{q}+{(2\mathrm{x}-3)}^2\cosh \mathrm{q}$$$$=2\sinh ({\mathrm{x}}^2-3\mathrm{x}+2)+{(2\mathrm{x}-3)}^2\cosh ({\mathrm{x}}^2-3\mathrm{x}+1)$$

Answered by mathmax by abdo last updated on 26/Jun/21

$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{ch}({\mathrm{x}}^2-3\mathrm{x}+1)=\frac{{\mathrm{e}}^{{\mathrm{x}}^2-3\mathrm{x}+1}+{\mathrm{e}}^{-{\mathrm{x}}^2+3\mathrm{x}-1}}{2}\Rightarrow$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{2}((2\mathrm{x}-3){\mathrm{e}}^{{\mathrm{x}}^2-3\mathrm{x}+1}+(-2\mathrm{x}+3){\mathrm{e}}^{-{\mathrm{x}}^2+3\mathrm{x}-1})\Rightarrow$$$${\mathrm{y}}^{\left(2\right)}\left(\mathrm{x}\right)=\frac{1}{2}\{{2\mathrm{e}}^{{\mathrm{x}}^2-3\mathrm{x}+1}+{(2\mathrm{x}-3)}^2{\mathrm{e}}^{{\mathrm{x}}^2-3\mathrm{x}+1}-{2\mathrm{e}}^{-{\mathrm{x}}^2+3\mathrm{x}-1}+{(-2\mathrm{x}+3)}^2{\mathrm{e}}^{-{\mathrm{x}}^2+3\mathrm{x}-1}\}$$$$=\frac{1}{2}\{(11+{4\mathrm{x}}^2-12\mathrm{x}){\mathrm{e}}^{{\mathrm{x}}^2-3\mathrm{x}+1}+(7+{4\mathrm{x}}^2-12\mathrm{x}){\mathrm{e}}^{-{\mathrm{x}}^2+3\mathrm{x}-1}\}$$

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