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Pls HELP on differentiation - HSN forum

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Pls HELP on differentiation


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#1 cK_hoangan

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Posted 30 November 2007 - 08:26 PM

1. y = cos (3x - pi/6)- cos2 ( tan (x) ) - sin (pi/3) , dy/dx = ?

2. y= (ln2 (x) + lnx + 1)^1/2 , dy/dx = ?

3. y= ( tan (3x) - cotg (3x) )^1/2, dy/dx = ?

#2 Steve

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Posted 03 December 2007 - 10:18 AM

Hi there,

These are all exercises in using the chain rule (over and over again). For the first one:



\begin{align*}
  y =  & \cos \left( {3x - \frac{\pi }
{6}} \right) - \left( {\cos \left( {\tan x} \right)} \right)^2  - \underbrace {\sin \left( {\frac{\pi }
{3}} \right)}_{{\text{constant}}} \\ 
  \frac{{dy}}
{{dx}} =  &  - \sin \left( {3x - \frac{\pi }
{6}} \right) \times \frac{d}
{{dx}}\left( {3x - \frac{\pi }
{6}} \right) - 2\left( {\cos \left( {\tan x} \right)} \right) \times \frac{d}
{{dx}}\left( {\cos \left( {\tan x} \right)} \right) \\ 
   =  &  - \sin \left( {3x - \frac{\pi }
{6}} \right) \times 3 - 2\left( {\cos \left( {\tan x} \right)} \right) \times  - \sin \left( {\tan x} \right) \times \underbrace {\frac{d}
{{dx}}\left( {\tan x} \right)}_{{\text{chain rule again}}} \\ 
   =  &  - 3\sin \left( {3x - \frac{\pi }
{6}} \right) + 2\cos \left( {\tan x} \right)\sin \left( {\tan x} \right)\sec ^2 x
\end{align*}

I hope this helps (and I haven't made any mistakes!). Do you get the idea now?
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