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# forgotten how to differentiate... watch

1. In uni now, forgotten my alevel maths. could anyone help with this simple question pls.

differentiate y=ln(cosx)

2. y' = -sinx/cosx = -tanx
3. By using the Chain rule:

(dy/dx)=((du)/(dx))((dy)/(du))

Let u=cosx => (du/dx)=(-(sinx))

y=lnu => (dy)/(du)=1/u

Therefore:

(dy/dx)=(1/u)(-(sinx))=(1/cos)(-(sinx))=-tanx

Newton.
4. (Original post by Newton)
By using the Chain rule:

(dy/dx)=((du)/(dx))((dy)/(du))

Let u=cosx => (du/dx)=(-(sinx))

y=lnu => (dy)/(du)=1/u

Therefore:

(dy/dx)=(1/u)(-(sinx))=(1/cos)(-(sinx))=-tanx

Newton.
just remember d/dx [ln (u)] = 1/u * du/dx
in thiis case it is 1/cos x * -sin x - -tan x

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Updated: January 19, 2005
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