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

Name: 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
Uploaded: 2022-02-23T17:22:57+00:00
Description: 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

Question

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

Jeffery Autrey · Accepted Answer

dydx=1cos⁡y−sin⁡y.Take the derivative of both sides using one of the derivative rules:d2ydx2=(1)′(cos⁡y−sin⁡y)−(1)(cos⁡y−sin⁡y)′(cos⁡y−sin⁡y)2=?,or, (x−1)′=−x−2⇒d2ydx2=−1(cos⁡y−sin⁡y)2⋅(cos⁡y−sin⁡y)′=?Above are the beginnings to (i) the quotient rule and (ii) the power rule and chain rules.

Marcus Herman · Accepted Answer

Just differentiate both sides of dydx=(cos⁡y−sin⁡y)−1 with respect to x. This leads to:  d2ydx2=ddxdydx  =ddx(cos⁡y−sin⁡y)−1=−(cos⁡y−sin⁡y)−2⋅[(−sin⁡y)dydx−(cos⁡y)dydx].

RizerMix · Accepted Answer

x=sin⁡y+cos⁡y first derivative by implicit diffferentiation;1=cos⁡y(y′)−sin⁡y(y′)1=y′(cos⁡y−sin⁡y)y′=1/(cos⁡y−sin⁡y) second derivetive y″=−(−sin⁡y(y′)−cos⁡y(y′))/(cos⁡y−sin⁡y)2y″=(y′sin⁡y+y′cos⁡y)/(cos⁡y−sin⁡y)2 subsitute the value of y&#039; we hwve;y″=(1/(cos⁡y−sin⁡y)(sin⁡y+cos⁡y))/(cos⁡y−sin⁡y)2y″=(sin⁡y+cos⁡y)/(cos⁡y−sin⁡y)2 simplifying this we have;y″=(sin⁡y+cos⁡y)/(−sin⁡2y) or y″=−(sin⁡y+cos⁡y)/2sin⁡ycos⁡y y″=−sin⁡y/2sin⁡ycos⁡y−cos⁡y/2sin⁡ycos⁡yy″=−sec⁡y/2−csc⁡y/2