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Question Number 109724 by bobhans last updated on 25/Aug/20

$$y=\sqrt{x+\sqrt{x+\sqrt{x}}}\Rightarrow \frac{dy}{dx}?$$

Answered by ajfour last updated on 25/Aug/20

$${(y^2-x)}^2=x+\sqrt{x}$$$$2(y^2-x)(2y\frac{dy}{dx}-1)=1+\frac{1}{2\sqrt{x}}$$$$\frac{dy}{dx}=\frac{1}{2y}\{1+\frac{1+\frac{1}{2\sqrt{x}}}{2(y^2-x)}\}$$$$\frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\{1+\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}\}$$

Answered by bemath last updated on 25/Aug/20

Answered by 1549442205PVT last updated on 25/Aug/20

$$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}.\mathrm{Squaring}\mathrm{two}\mathrm{sides}\mathrm{we}\mathrm{get}$$$${\mathrm{y}}^2=\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}.\mathrm{Derivative}\mathrm{two}\mathrm{sides}$$$$\mathrm{by}\mathrm{x}\mathrm{we}\mathrm{get}2\mathrm{y}.{\mathrm{y}}^{\prime }=1+\frac{{(\mathrm{x}+\sqrt{\mathrm{x}})}^{\prime }}{2\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}$$$$2\mathrm{y}.{\mathrm{y}}^{\prime }=1+\frac{1+\frac{1}{2\sqrt{\mathrm{x}}}}{2\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}=1+\frac{2\sqrt{\mathrm{x}}+1}{4\sqrt{\mathrm{x}}.\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}$$$$=\frac{4\sqrt{{\mathrm{x}}^2+\mathrm{x}\sqrt{\mathrm{x}}}+2\sqrt{\mathrm{x}}+1}{4\sqrt{{\mathrm{x}}^2+\mathrm{x}\sqrt{\mathrm{x}}}}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\sqrt{{\mathrm{x}}^2+\mathrm{x}\sqrt{\mathrm{x}}}+2\sqrt{\mathrm{x}}+1}{4\sqrt{{\mathrm{x}}^2+\mathrm{x}\sqrt{\mathrm{x}}}}/2\mathrm{y}$$$$=\frac{4\sqrt{{\mathbf{x}}^2+\mathbf{x}\sqrt{\mathbf{x}}}+2\sqrt{\mathbf{x}}+1}{2\sqrt{\mathbf{x}+\sqrt{\mathbf{x}+\sqrt{\mathbf{x}}}.}\left(4\sqrt{{\mathbf{x}}^2+\mathbf{x}\sqrt{\mathbf{x}}}\right)}$$$$\mathbf{Check}\mathbf{by}\mathbf{calculator}$$$$\frac{\mathbf{d}}{\mathbf{dx}}\left(\sqrt{\mathbf{x}+\sqrt{\mathbf{x}+\sqrt{\mathbf{x}}}}\right){\mid }_{\mathbf{x}=1}=0.4924558242$$$$=\frac{4\sqrt{{\mathbf{x}}^2+\mathbf{x}\sqrt{\mathbf{x}}}+2\sqrt{\mathbf{x}}+1}{2\sqrt{\mathbf{x}+\sqrt{\mathbf{x}+\sqrt{\mathbf{x}}}.}\left(4\sqrt{{\mathbf{x}}^2+\mathbf{x}\sqrt{\mathbf{x}}}\right)}{\mid }_{\mathrm{x}=1}$$

Answered by mathmax by abdo last updated on 25/Aug/20

$$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}\Rightarrow {\mathrm{y}}^2=\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}\Rightarrow {({\mathrm{y}}^2\left(\mathrm{x}\right)-\mathrm{x})}^2=\mathrm{x}+\sqrt{\mathrm{x}}\Rightarrow$$$$2(2\mathrm{y}\left(\mathrm{x}\right){\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)-1)({\mathrm{y}}^2\left(\mathrm{x}\right)-1)=1+\frac{1}{2\sqrt{\mathrm{x}}}\Rightarrow$$$$(4\mathrm{y}\left(\mathrm{x}\right){\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)-2)(\mathrm{x}-1+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}})=\frac{2\sqrt{\mathrm{x}}+1}{2\sqrt{\mathrm{x}}}\Rightarrow$$$$4\mathrm{y}\left(\mathrm{x}\right){\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{2\sqrt{\mathrm{x}}+1}{2\sqrt{\mathrm{x}}(\mathrm{x}-1+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}})}+2\Rightarrow$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{4\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}\{\frac{2\sqrt{\mathrm{x}}+1}{2\sqrt{\mathrm{x}}(\mathrm{x}-1+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}+2\}$$

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