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Question Number 168276 by Florian last updated on 07/Apr/22

    y = (√(x+(√(x+(√x)))))      y′ =

$$\:\:\:\:{y}\:=\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}} \\ $$$$\:\:\:\:{y}'\:=\: \\ $$$$\:\:\:\:\: \\ $$

Commented by cortano1 last updated on 07/Apr/22

 y′=((1+((1+(1/(2(√x))))/(2(√(x+(√x))))))/(2(√(x+(√(x+(√x)))))))

$$\:{y}'=\frac{\mathrm{1}+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}}}}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}} \\ $$

Commented by Florian last updated on 08/Apr/22

???

$$??? \\ $$

Answered by greogoury55 last updated on 07/Apr/22

 y=(√(x+(√(x+(√x)))))   y^2 −x = (√(x+(√x)))    (y^2 −x)^2 −x=(√x)   (d/dx) ((y^2 −x)^2 −x)= (d/dx)((√x) )   2(y^2 −x)(2yy′−1)−1 = (1/(2(√x)))  ⇒2(y^2 −x)(2yy′−1)=((1+2(√x))/(2(√x)))  ⇒2yy′−1= ((1+2(√x))/(4(√x) (√(x+(√x))))) =((1+2(√x))/(4(√(x^2 +x(√x)))))  ⇒2yy′= ((1+2(√x)+4(√(x^2 +x(√x))))/(4(√(x^2 +x(√x)))))  ⇒y′ = ((1+2(√x)+4(√(x^2 +x(√x))))/(8(√(x^2 +x(√x))) (√(x+(√(x+(√x)))))))  ⇒y′= ((1+2(√x)+4(√(x^2 +x(√x))))/(8(√((x^2 +x(√x))(x+(√(x+(√x))))))))

$$\:{y}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}} \\ $$$$\:{y}^{\mathrm{2}} −{x}\:=\:\sqrt{{x}+\sqrt{{x}}}\: \\ $$$$\:\left({y}^{\mathrm{2}} −{x}\right)^{\mathrm{2}} −{x}=\sqrt{{x}} \\ $$$$\:\frac{{d}}{{dx}}\:\left(\left({y}^{\mathrm{2}} −{x}\right)^{\mathrm{2}} −{x}\right)=\:\frac{{d}}{{dx}}\left(\sqrt{{x}}\:\right) \\ $$$$\:\mathrm{2}\left({y}^{\mathrm{2}} −{x}\right)\left(\mathrm{2}{yy}'−\mathrm{1}\right)−\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\Rightarrow\mathrm{2}\left({y}^{\mathrm{2}} −{x}\right)\left(\mathrm{2}{yy}'−\mathrm{1}\right)=\frac{\mathrm{1}+\mathrm{2}\sqrt{{x}}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\Rightarrow\mathrm{2}{yy}'−\mathrm{1}=\:\frac{\mathrm{1}+\mathrm{2}\sqrt{{x}}}{\mathrm{4}\sqrt{{x}}\:\sqrt{{x}+\sqrt{{x}}}}\:=\frac{\mathrm{1}+\mathrm{2}\sqrt{{x}}}{\mathrm{4}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}}}} \\ $$$$\Rightarrow\mathrm{2}{yy}'=\:\frac{\mathrm{1}+\mathrm{2}\sqrt{{x}}+\mathrm{4}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}}}}{\mathrm{4}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}}}} \\ $$$$\Rightarrow{y}'\:=\:\frac{\mathrm{1}+\mathrm{2}\sqrt{{x}}+\mathrm{4}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}}}}{\mathrm{8}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}}}\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}} \\ $$$$\Rightarrow{y}'=\:\frac{\mathrm{1}+\mathrm{2}\sqrt{{x}}+\mathrm{4}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}}}}{\mathrm{8}\sqrt{\left({x}^{\mathrm{2}} +{x}\sqrt{{x}}\right)\left({x}+\sqrt{{x}+\sqrt{{x}}}\right)}}\: \\ $$$$ \\ $$

Commented by Florian last updated on 07/Apr/22

Very Good!

$${Very}\:{Good}! \\ $$

Answered by Mathspace last updated on 07/Apr/22

y^2 =x+(√(x+(√x)))  ⇒(y^2 −x)^2 =x+(√x) ⇒  2(2yy^′ −1)(y^2 −x)=1+(1/(2(√x))) ⇒  4yy^′ =((2(√x)+1)/(2(√x)(y^2 −x)))=((2(√x)+1)/(2(√x)(√(x+(√x)))))+2  ⇒y^′ =(1/(4y)){((3(√x)+1)/(2(√(x^2 +x(√x))))) +2}

$${y}^{\mathrm{2}} ={x}+\sqrt{{x}+\sqrt{{x}}} \\ $$$$\Rightarrow\left({y}^{\mathrm{2}} −{x}\right)^{\mathrm{2}} ={x}+\sqrt{{x}}\:\Rightarrow \\ $$$$\mathrm{2}\left(\mathrm{2}{yy}^{'} −\mathrm{1}\right)\left({y}^{\mathrm{2}} −{x}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\:\Rightarrow \\ $$$$\mathrm{4}{yy}^{'} =\frac{\mathrm{2}\sqrt{{x}}+\mathrm{1}}{\mathrm{2}\sqrt{{x}}\left({y}^{\mathrm{2}} −{x}\right)}=\frac{\mathrm{2}\sqrt{{x}}+\mathrm{1}}{\mathrm{2}\sqrt{{x}}\sqrt{{x}+\sqrt{{x}}}}+\mathrm{2} \\ $$$$\Rightarrow{y}^{'} =\frac{\mathrm{1}}{\mathrm{4}{y}}\left\{\frac{\mathrm{3}\sqrt{{x}}+\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{{x}}}}\:+\mathrm{2}\right\} \\ $$

Answered by Florian last updated on 07/Apr/22

     y =(√(x+(√(x+(√x)))))         y′=((√(x+(√(x+(√x))))))′             →g_1 =x+(√(x+(√x)))         y′=((√g_1 ))′(x+(√(x+(√x))))′         y′=(1/(2(√g_1 )))((x)′+((√(x+(√x))))′)     →g_2 =x+(√x)         y′=(1/(2(√g_1 )))(1+((√g_2 ))′(x+(√x))′         y′=(1/(2(√g_1 )))(1+(1/(2(√g_2 )))(1+(1/( 2(√x))))         y′=(1/(2(√g_1 )))(1+(1/(2(√(x+(√x)))))(1+(1/(2(√x)))))         y′=(1/(2(√g_1 )))(1+((2(√(x+1)))/(4(√(x^2 +(√x)x)))))         y′=(1/(2(√(x+(√(x+(√x)))))))(1+((2(√(x+1)))/(4(√(x^2 +(√x)x)))))         y′=(1/(2(√(x+(√(x(√x)))))))×((4(√(x^2 +(√x)x))+2(√x)+1)/(4(√(x^2 +(√x)x))))         y′=((4(√(x^2 +(√x)x))+2(√x)+1)/(8(√((x+(√(x+(√x))))(x^2 +(√x)x)))))

$$\:\:\:\:\:{y}\:=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}} \\ $$$$\:\:\:\:\:\:\:{y}'=\left(\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}\right)'\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrow{g}_{\mathrm{1}} ={x}+\sqrt{{x}+\sqrt{{x}}} \\ $$$$\:\:\:\:\:\:\:{y}'=\left(\sqrt{{g}_{\mathrm{1}} }\right)'\left({x}+\sqrt{{x}+\sqrt{{x}}}\right)' \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{g}_{\mathrm{1}} }}\left(\left({x}\right)'+\left(\sqrt{{x}+\sqrt{{x}}}\right)'\right)\:\:\:\:\:\rightarrow{g}_{\mathrm{2}} ={x}+\sqrt{{x}} \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{g}_{\mathrm{1}} }}\left(\mathrm{1}+\left(\sqrt{{g}_{\mathrm{2}} }\right)'\left({x}+\sqrt{{x}}\right)'\right. \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{g}_{\mathrm{1}} }}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{g}_{\mathrm{2}} }}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{x}}}\right)\right. \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{g}_{\mathrm{1}} }}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}}}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)\right) \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{g}_{\mathrm{1}} }}\left(\mathrm{1}+\frac{\mathrm{2}\sqrt{{x}+\mathrm{1}}}{\mathrm{4}\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}}{x}}}\right) \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}\left(\mathrm{1}+\frac{\mathrm{2}\sqrt{{x}+\mathrm{1}}}{\mathrm{4}\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}}{x}}}\right) \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}\sqrt{{x}}}}}×\frac{\mathrm{4}\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}}{x}}+\mathrm{2}\sqrt{{x}}+\mathrm{1}}{\mathrm{4}\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}}{x}}} \\ $$$$\:\:\:\:\:\:\:{y}'=\frac{\mathrm{4}\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}}{x}}+\mathrm{2}\sqrt{{x}}+\mathrm{1}}{\mathrm{8}\sqrt{\left({x}+\sqrt{{x}+\sqrt{{x}}}\right)\left({x}^{\mathrm{2}} +\sqrt{{x}}{x}\right)}} \\ $$

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