1. ## [SOLVED] Differentiate

I want to differentiate this $\displaystyle 2e^{2x}(Acos(x)-Bsin(x))+e^{2x}(-Asin(x)+Bcos(x))$

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

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

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

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

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

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

I need to differentiate this when I try it got really messy and could not complete it.
Take out a common factor of $\displaystyle e^{2x}$ and you get

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

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

Now differentiate using the product rule

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

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

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

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

Now differentiate using the product rule

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

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

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

Would u=$\displaystyle 2e^{2x}$ and v=$\displaystyle cos{x} - sin{x}$
No. Let $\displaystyle u = e^{2x}$ and $\displaystyle v =$ everything in the square brackets.

So the derivative is $\displaystyle \frac{du}{dx}\times v + u\times\frac{dv}{dx}$.

5. I differentiated the first part $\displaystyle 2e^{2x}[(2A + B)\cos{x} - (A + 2B)\sin{x}]$

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

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

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

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

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