Second derivative using implicit differentiation with respect to x of x=sin⁡y+cos⁡y 1=cos⁡ydydx−sin⁡ydydx 1=dydx(cos⁡y−sin⁡y) dydx=1cos⁡y−sin⁡y

burkinaval1b

burkinaval1b

Answered

2022-01-21

Second derivative using implicit differentiation with respect to x of x=siny+cosy
1=cosydydxsinydydx
1=dydx(cosysiny)
dydx=1cosysiny

Answer & Explanation

Jeffery Autrey

Jeffery Autrey

Expert

2022-01-21Added 35 answers

dydx=1cosysiny.
Take the derivative of both sides using one of the derivative rules:
d2ydx2=(1)(cosysiny)(1)(cosysiny)(cosysiny)2=?,
or, (x1)=x2d2ydx2=1(cosysiny)2(cosysiny)=?
Above are the beginnings to (i) the quotient rule and (ii) the power rule and chain rules.
Marcus Herman

Marcus Herman

Expert

2022-01-22Added 41 answers

Just differentiate both sides of dydx=(cosysiny)1 with respect to x. This leads to:
d2ydx2=ddxdydx
=ddx(cosysiny)1=(cosysiny)2[(siny)dydx(cosy)dydx].
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

x=siny+cosy first derivative by implicit diffferentiation;1=cosy(y)siny(y)1=y(cosysiny)y=1/(cosysiny) second derivetive y=(siny(y)cosy(y))/(cosysiny)2y=(ysiny+ycosy)/(cosysiny)2 subsitute the value of y' we hwve;y=(1/(cosysiny)(siny+cosy))/(cosysiny)2y=(siny+cosy)/(cosysiny)2 simplifying this we have;y=(siny+cosy)/(sin2y) or y=(siny+cosy)/2sinycosy y=siny/2sinycosycosy/2sinycosyy=secy/2cscy/2

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