# Confusion on a differentiation questionWatch

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#1
for question 8, I can't seem to prove it. Attachment 593372593374

Thanks
0
2 years ago
#2
(Original post by coconut64)
for question 8, I can't seem to prove it. Attachment 593372593374

Thanks
x = sec y

d/dx(x) = d/dy(sec y))

1 = dy/dx sec y tan y

dy /dx = 1 / (sec y tan y)

tan^2 y + 1= sec^2 y
tan y = √(sec^2y - 1)

dy/dx = 1 / [x √(x^2 - 1)]

hope that helps
1
#3
(Original post by maruchan)
x = sec y

d/dx(x) = d/dy(sec y))

1 = dy/dx sec y tan y

dy /dx = 1 / (sec y tan y)

tan^2 y + 1= sec^2 y
tan y = √(sec^2y - 1)

dy/dx = 1 / [x √(x^2 - 1)]

hope that helps
Thanks, that is a very detailed explanation ! however do u know that's implicit differentiation ? I thought it was a normal differentiation with subsitutuion ..
0
2 years ago
#4
(Original post by coconut64)
for question 8, I can't seem to prove it.
Thanks

What you worked out on the left is dx/dy, NOT dy/dx.

Using dy/dx = 1/(dx/dy) would put you back on the right track.
0
#5
(Original post by ghostwalker)

What you worked out on the left is dx/dy, NOT dy/dx.

Using dy/dx = 1/(dx/dy) would put you back on the right track.
Thanks
0
2 years ago
#6
(Original post by coconut64)
Thanks, that is a very detailed explanation ! however do u know that's implicit differentiation ? I thought it was a normal differentiation with subsitutuion ..

Yes i just used implicit differentiation because i found it easier to do it that way. As it says x=secy i think it's best to use implicit differentiation
0
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