# Simple differentiation problem, brainfart

1. Jul 26, 2009

### James889

Hai,

I have the easiest problem but im stuck at the last step, simplification.

$$\frac{cos3x+sin3x}{cos3x-sin3x}$$

\begin{aligned} F(x) = cos(3x)+sin(3x)\\ F'(x) = -3sin(3x)+3cos(3x) \\ G(x) = cos(3x)-sin(3x) \\ G'(x) = -3sin(3x)-3cos(3x)\end{aligned}

This gives $$\frac{(-3sin(3x)+3cos(3x))*(cos(3x)-sin(3x))+(cos(3x)+sin(3x))*(cos(3x)-sin(3x))}{(cos(3x)-sin(3x))^2}$$

But how would i simplify this?

2. Jul 26, 2009

### djeitnstine

Expand the quadratics and look for identities. Already I can find sin^2 + cos^2 identities on both the top and bottom.

3. Jul 26, 2009

### AUMathTutor

Also, it will help to remember that sin(u)cos(u) = (1/2)sin(2u)

4. Jul 26, 2009

### James889

Okay, after some more trying:

I was able to reduce the first part$$(-3sin(3x)+3cos(3x))(cos(3x)-sin(3x))$$
Down to:
$$-3sin(3x)*cos(3x) + 3 -3cos(3x)*sin(3x)$$

And the second part:$$(cos(3x)+sin(3x))*(cos(3x)-sin(3x))$$
Down to:
$$cos(3x)^{2}-sin(3x)^{2}-2sin(3x)cos(3x)$$

Is that any better?

5. Jul 26, 2009

### djeitnstine

Take note of AUMathtutor's comment

6. Jul 27, 2009

### AUMathTutor

Also, naturally,

cos(u)^{2} - sin(u)^{2} = cos(2u)