# [SOLVED] Differentiate

• Mar 1st 2009, 03:19 PM
ronaldo_07
[SOLVED] Differentiate
I want to differentiate this $2e^{2x}(Acos(x)-Bsin(x))+e^{2x}(-Asin(x)+Bcos(x))$

I have put the a terms together to make it simpler am i right in doing so?

$Ae^{2x}(2cos(x) - sin(x)))+ Be^{2x}(2sin(x) + cos(x))$

I need to differentiate this when I try it got really messy and could not complete it.
• Mar 1st 2009, 03:35 PM
Prove It
Quote:

Originally Posted by ronaldo_07
I want to differentiate this $2e^{2x}(Acos(x)-Bsin(x))+e^{2x}(-Asin(x)+Bcos(x))$

I have put the a terms together to make it simpler am i right in doing so?

$Ae^{2x}(2cos(x) - sin(x)))+ Be^{2x}(2sin(x) + cos(x))$

I need to differentiate this when I try it got really messy and could not complete it.

Take out a common factor of $e^{2x}$ and you get

$f(x) = e^{2x}(2A\cos{x} - 2B\sin{x} - A\sin{x} + B\cos{x})$

$f(x) = e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]$

Now differentiate using the product rule

$f'(x) = 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}] - e^{2x}[(2A + B)\sin{x} + (A + 2B)\cos{x}]$.
• Mar 1st 2009, 03:50 PM
ronaldo_07
Quote:

Originally Posted by Prove It
Take out a common factor of $e^{2x}$ and you get

$f(x) = e^{2x}(2A\cos{x} - 2B\sin{x} - A\sin{x} + B\cos{x})$

$f(x) = e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]$

Now differentiate using the product rule

$f'(x) = 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}] - e^{2x}[(2A + B)\sin{x} + (A + 2B)\cos{x}]$.

for the first part: $2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]$

Would u= $2e^{2x}$ and v= $cos{x} - sin{x}$
• Mar 1st 2009, 03:53 PM
Prove It
Quote:

Originally Posted by ronaldo_07
for the first part: $2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]$

Would u= $2e^{2x}$ and v= $cos{x} - sin{x}$

No. Let $u = e^{2x}$ and $v =$ everything in the square brackets.

So the derivative is $\frac{du}{dx}\times v + u\times\frac{dv}{dx}$.
• Mar 1st 2009, 04:23 PM
ronaldo_07
I differentiated the first part $2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]$

and got $2e^{2x}[(2A+B)cos(x)-(A+2B)sin(x)]-2e^{2x}[(2A+B)sin(x)+(A+2B)cos(x)]$

is this correct? then i differentiate the other part and put them together?
• Mar 1st 2009, 08:56 PM
Prove It
Quote:

Originally Posted by ronaldo_07
I differentiated the first part $2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]$

and got $2e^{2x}[(2A+B)cos(x)-(A+2B)sin(x)]-2e^{2x}[(2A+B)sin(x)+(A+2B)cos(x)]$

is this correct? then i differentiate the other part and put them together?