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Question Number 66101 by Rio Michael last updated on 09/Aug/19

$$find\frac{dy}{dx}wheny=x^{2}ln\left(3x\right)$$$$Giventhatxsinx-y^2=0showthat{y}^2=2cosx-2{\left(\frac{dy}{dx}\right)}^2-2y\frac{d^{2}y}{{dx}^2}$$

Commented by Prithwish sen last updated on 09/Aug/19

$$\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{xln}3\mathrm{x}+\mathrm{x}$$$${\mathrm{y}}^2=\mathrm{xsinx}$$$$2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{sinx}+\mathrm{xcosx}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{sinx}+\mathrm{xcosx}}{2\sqrt{\mathrm{xsinx}}}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{1}{2}.\frac{\mathrm{y}(2\mathrm{cosx}-\mathrm{xsinx})-\frac{{(\mathrm{sinx}+\mathrm{xcosx})}^2}{2\mathrm{y}}}{{\mathrm{y}}^2}$$$${4\mathrm{y}}^3\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={2\mathrm{y}}^2(2\mathrm{cosx}-{\mathrm{y}}^2)-{\left(2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2$$$$2\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2\mathrm{cosx}-{\mathrm{y}}^2-2{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2$$$$\mathrm{proved}.$$

Answered by GordonYeeman last updated on 09/Aug/19

$$\frac{dy}{dx}=2xln\left(3x\right)+x^2\frac{1}{3x}3=x(2ln\left(3x\right)+1)$$

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