Use-implicit-differentiation-to-find-d-2-y-dx-2-for-siny-x-

Question Number 184040 by pete last updated on 02/Jan/23

$$\mathrm{Use}\:\mathrm{implicit}\:\mathrm{differentiation}\:\mathrm{to}\:\mathrm{find}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} } \\ $$$$\mathrm{for}\:\mathrm{sin}{y}\:=\:{x} \\ $$

Answered by cortano2 last updated on 02/Jan/23

$${y}'\mathrm{cos}\:{y}=\mathrm{1} \\ $$$${y}''\left(−\mathrm{sin}\:{y}\right)=\mathrm{0} \\ $$$${y}''=\mathrm{0} \\ $$

Answered by Yhusuph last updated on 02/Jan/23

(dy/dx)cosy = 1 (dy/dx)=(1/( (√(1−x^2 )))) (d^2 y/dx^2 ) = ((−(−2(1−x^2 )^(−(1/2)) ))/(1−x^2 )) (d^2 y/dx^2 ) = (2/((1−x^2 )^(3/2) )) (d^2 y/dx^2 ) = (2/((^2 (√(1−x^2 )))^3 )) ∂enken Last βorn Mentor : Proffyemphy

$$ \\ $$$$\frac{{dy}}{{dx}}{cosy}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{−\left(−\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}}{\left(^{\mathrm{2}} \sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }\right)^{\mathrm{3}} } \\ $$$$\partial{enken}\:{Last}\:\beta{orn} \\ $$$${Mentor}\::\:{Proffyemphy} \\ $$$$ \\ $$

Commented by Frix last updated on 02/Jan/23

This cannot be true. y=arcsin x ⇒ (d^2 y/dx^2 )=(x/((1−x^2 )^(3/2) ))

$$\mathrm{This}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{true}. \\ $$$${y}=\mathrm{arcsin}\:{x}\:\Rightarrow\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$

Answered by mr W last updated on 03/Jan/23

sin y=x y′ cos y=1 y′′ cos y−(y′)^2 sin y=0 y′′=(y′)^2 ((sin y)/(cos y))=((tan y)/(cos^2 y))=((sin y)/(cos^3 y))=((sin y)/((1−sin^2 y)^(3/2) )) y′′=(x/((1−x^2 )^(3/2) ))

$$\mathrm{sin}\:{y}={x} \\ $$$${y}'\:\mathrm{cos}\:{y}=\mathrm{1} \\ $$$${y}''\:\mathrm{cos}\:{y}−\left({y}'\right)^{\mathrm{2}} \:\mathrm{sin}\:{y}=\mathrm{0} \\ $$$${y}''=\left({y}'\right)^{\mathrm{2}} \:\frac{\mathrm{sin}\:{y}}{\mathrm{cos}\:{y}}=\frac{\mathrm{tan}\:{y}}{\mathrm{cos}^{\mathrm{2}} \:{y}}=\frac{\mathrm{sin}\:{y}}{\mathrm{cos}^{\mathrm{3}} \:{y}}=\frac{\mathrm{sin}\:{y}}{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{y}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${y}''=\frac{{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$

Commented by pete last updated on 02/Jan/23

Thank you Mr. W, but the answer given is: (d^2 y/dx^2 ) = sec^2 ytany

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Mr}.\:\mathrm{W},\:\mathrm{but}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{given} \\ $$$$\mathrm{is}:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\mathrm{sec}^{\mathrm{2}} \mathrm{ytany} \\ $$

Commented by cortano1 last updated on 03/Jan/23

from sin y = x we get { ((sec y=(1/( (√(1−x^2 )))))),((tan y=(x/( (√(1−x^2 )))))) :} now y′′= (x/((1−x^2 )(√(1−x^2 )))) = [(1/( (√(1−x^2 )))) ]^2 .(x/( (√(1−x^2 )))) = sec^2 y tan y

$$\:{from}\:\mathrm{sin}\:{y}\:=\:{x}\:{we}\:{get}\:\begin{cases}{\mathrm{sec}\:{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}\\{\mathrm{tan}\:{y}=\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}\end{cases} \\ $$$${now}\:{y}''=\:\frac{{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\: \\ $$$$\:\:\:\:=\:\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\right]^{\mathrm{2}} .\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:\mathrm{sec}\:^{\mathrm{2}} {y}\:\mathrm{tan}\:{y}\: \\ $$

Commented by mr W last updated on 03/Jan/23

you can express y′′ in terms of y or in terms of x. you see both above. y′′=(y′)^2 ((sin y)/(cos y))=((tan y)/(cos^2 y))=((sin y)/(cos^3 y))=((sin y)/((1−sin^2 y)^(3/2) )) y′′=(x/((1−x^2 )^(3/2) ))

$${you}\:{can}\:{express}\:{y}''\:{in}\:{terms}\:{of}\:{y}\:{or} \\ $$$${in}\:{terms}\:{of}\:{x}.\:{you}\:{see}\:{both}\:{above}. \\ $$$${y}''=\left({y}'\right)^{\mathrm{2}} \:\frac{\mathrm{sin}\:{y}}{\mathrm{cos}\:{y}}=\frac{\mathrm{tan}\:{y}}{\mathrm{cos}^{\mathrm{2}} \:{y}}=\frac{\mathrm{sin}\:{y}}{\mathrm{cos}^{\mathrm{3}} \:{y}}=\frac{\mathrm{sin}\:{y}}{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{y}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${y}''=\frac{{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$

Commented by pete last updated on 04/Jan/23

$$\mathrm{Thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

Answered by manxsol last updated on 02/Jan/23

$$ \\ $$$$ \\ $$

Commented by mr W last updated on 03/Jan/23

$${maybe}\:{not}\:{all}\:{people}\:{can}\:{read}\:{your} \\ $$$${solution}\:{properly}.\:{this}\:{is}\:{how}\:{your} \\ $$$${post}\:{looks}\:{like}: \\ $$

Commented by mr W last updated on 03/Jan/23