# Stationary point...

• June 9th 2010, 12:54 AM
Punch
Stationary point...
Find the coordinates of the stationary point of the curve $y=e^xcosx$ for 0 is less than or equal to x and x is less than or equal to $/pi$ . Leave your answer in exact form.
• June 9th 2010, 12:58 AM
mr fantastic
Quote:

Originally Posted by Punch
Find the coordinates of the stationary point of the curve $y=e^xcosx$ for 0 is less than or equal to x and x is less than or equal to x . Leave your answer in exact form.

First get dy/dx (use the product rule). Can you do that? Please show all your work and say where you're stuck.
• June 9th 2010, 12:59 AM
Punch
$
dy/dx=-e^xsinx
$

am stucked at the part solving for x
• June 9th 2010, 01:03 AM
mr fantastic
Quote:

Originally Posted by Punch
$
dy/dx=-e^xsinx
$

[snip]

Wrong. Have you been taught the product rule? If not, then attempting this question is currently a futile exercise for you. If you have, then I suggest you go to your class notes or textbook and review it.
• June 9th 2010, 01:09 AM
Punch
$\frac{dy}{dx} = e^xcosx-e^xsinx$
$= e^x[cosx-sinx]$

now, $e^x[cosx-sinx]=0$ stucked =="
• June 9th 2010, 01:12 AM
mr fantastic
Quote:

Originally Posted by Punch
$\frac{dy}{dx} = e^xcosx-e^xsinx$
$= e^x[cosx-sinx]$

Correct.

Now solve $\frac{dy}{dx} = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow 1 = \tan x$ over the given domain.

Then find the y-coordinate corresponding to the solutions for x found above.

If you need more help, please show all your work and say where you get stuck.
• June 9th 2010, 01:27 AM
Punch
i dont understand the part where u "threw" e^x away... wont u lose a solution? other than this everything else is fine :) thx
• June 9th 2010, 06:57 PM
mr fantastic
Quote:

Originally Posted by Punch
i dont understand the part where u "threw" e^x away... wont u lose a solution? other than this everything else is fine :) thx

Do you know what the graph of y = e^x looks like? Can e^x ever equal zero?