- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

View attachment 9064

that was the example but....

\begin{align*}

w(e^x,\cos x)&=\left|\begin{array}{rr}e^x&\cos{x} \\ e^x&-\cos{x} \\ \end{array}\right|\\

&=??\\

&=??

\end{align*}

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,886

View attachment 9064

that was the example but....

\begin{align*}

w(e^x,\cos x)&=\left|\begin{array}{rr}e^x&\cos{x} \\ e^x&-\cos{x} \\ \end{array}\right|\\

&=??\\

&=??

\end{align*}

- Admin
- #2

- Jan 26, 2012

- 4,198

But don't you need to calculate $w\left(e^{2x},\cos(2x)\right)?$ What do you get for that?

- Mar 1, 2012

- 849

\(\displaystyle w(e^x,\cos{x}) = \begin{vmatrix}but....

\begin{align*}

w(e^x,\cos x)&=\left|\begin{array}{rr}e^x&\cos{x} \\ e^x&-\cos{x} \\ \end{array}\right|\\

&=??\\

&=??

\end{align*}

e^x & \cos{x}\\

e^x & -\sin{x}

\end{vmatrix} = -e^x(\sin{x}+\cos{x})\)

- Thread starter
- #4

- Jan 31, 2012

- 2,886

Sorry everybody I think this thread went off the rails

my 2nd post was way off

so Ill post another new one of a similar problem

my 2nd post was way off

so Ill post another new one of a similar problem

Last edited: