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#1
given that

x= sec^2 y + tan y

show that dy/dx = (cos^2 y)/ (2tany +1)

kay I know I have to differentiate it then flip it over to make dy/dx isntead of dx/dy, but I can't differentiate it
can anyone help
0
13 years ago
#2

dx/dy = secytant*secy+ secytany*secy + sec^2 y
(using product rule on sec^2 y ie. (sec y*sec y))

dx/dy = (sec^2 y)(2tany + 1) (factorising out sec^2 y)

dx/dy = (2tany + 1)/ (cos^2 y) (using sec^2 y = (1/cos^2 y)

so dy/dx = (cos^2 y)/(2tany + 1) (using dy/dx = 1/(dx/dy))
0
#3
how did you get the first line??

does * mean multiplied by?
0
13 years ago
#4
(Original post by sarax)
given that

x= sec^2 y + tan y

show that dy/dx = (cos^2 y)/ (2tany +1)

kay I know I have to differentiate it then flip it over to make dy/dx isntead of dx/dy, but I can't differentiate it
can anyone help
x = (secy)^2 + tany
=> 1 = 2secy.secy.tany.dy/dx + sec^2y.dy/dx
=> 1 = dy/dx(2sec^2y.tany + sec^2y)
=> dy/dx = 1/(2sec^2y.tany + sec^2y)
=> dy/dx = 1/sec^2y.(2tany+1)
=> dy/dx = cos^2y/(2tany+1)
0
13 years ago
#5
(Original post by sarax)
how did you get the first line??

does * mean multiplied by?
yes
0
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