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Vahe10
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#1
Report Thread starter 17 years ago
#1
Ive just got theses two problems:
help wud be much appreciated:

1) e^2x + e^2y = xy

Find dy/dx.


2) y= 1/cosx

show that dy/dx = secx tanx

Thanx.
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Jesus09
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#2
Report 17 years ago
#2
(Original post by Vahe10)
Ive just got theses two problems:
help wud be much appreciated:

1) e^2x + e^2y = xy

Find dy/dx.


2) y= 1/cosx

show that dy/dx = secx tanx

Thanx.
wtf
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Ralfskini
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#3
Report 17 years ago
#3
(Original post by Vahe10)
Ive just got theses two problems:
help wud be much appreciated:

1) e^2x + e^2y = xy

Find dy/dx.


2) y= 1/cosx

show that dy/dx = secx tanx

Thanx.
e^2x + e^2y = xy

dy/dx = (y - 2e^2x)/(2e^2y - x)


2) dy/dx =-1cos^(-2)x * sinx = sinx/cos^2x = sinx/cosx* 1/cosx = secxtanx.
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Nylex
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Report 17 years ago
#4
(Original post by Vahe10)
Ive just got theses two problems:
help wud be much appreciated:

1) e^2x + e^2y = xy

Find dy/dx.


2) y= 1/cosx

show that dy/dx = secx tanx

Thanx.
1. Implicit differentiation

d/dx: 2e^2x + 2e^2y . dy/dx = y + xdy/dx

=> 2e^2y . dy/dx - xdy/dx = y - 2e^2x

=> dy/dx(2e^2y - x) = y - 2e^2x

=> dy/dx = (y - 2e^2x)/(2e^2y - x)

2. Quotient rule

d/dx (u/v) = (vdu/dx - udv/dx)/v^2

=> d/dx (1/cos x) = (cos x . 0 - 1 . -sin x)/cos^2 x

= sin x/cos^2 x

= (sinx/cos x) . 1/cos x

sin x/cos x = tan x, 1/cos x = sec x

=> d/dx = sec xtan x
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Ralfskini
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#5
Report 17 years ago
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(Original post by Nylex)
1. Implicit differentiation

d/dx: 2e^2x + 2e^2y . dy/dx = y + xdy/dx

=> 2e^2y . dy/dx - xdy/dx = y - 2e^2x

=> dy/dx(2e^2y - x) = y - 2e^2x

=> dy/dx = (y - 2e^2x)/(2e^2y - x)
same as me but with more work.
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Nylex
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#6
Report 17 years ago
#6
(Original post by Ralfskini)
same as me but with more work.
Yeah .
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elpaw
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#7
Report 17 years ago
#7
(Original post by Vahe10)
Ive just got theses two problems:
help wud be much appreciated:

1) e^2x + e^2y = xy

Find dy/dx.


2) y= 1/cosx

show that dy/dx = secx tanx

Thanx.
1) 2e^2x + 2e^2 dy/dx = y + x dy/dx

dy/dx (2e^2y - x) = y - 2e^2x

dy/dx = (y - 2 e^2x)/(2 e^2y -x)

2) y = sec x

dy/dx = secx tanx
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:P:P:P:P:P
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#8
Report 17 years ago
#8
(Original post by Ralfskini)
same as me but with more work.
How boastful was that post?!?
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