trig differential

**oxrigby** · November 6th 2008, 04:22 AM

Im doing this differential and used the chain/product rule up untuil i cannot simplify any further. Its quite messy and ive spent ages looking for a way to do it. The Q was differentiate with respect to x and simplify where possible

$sin^3xcos3x+cos^3xsin3x$

I got down to: $\frac{sin^3xsin3x}{3}+3sin^2xcos3xcosx+\frac{cos^3 xcos3x}{3}-sin3xcos^2xsinx$ a bit long dont you think?

**HallsofIvy** · November 6th 2008, 05:16 AM

Originally Posted by oxrigby

Im doing this differential and used the chain/product rule up untuil i cannot simplify any further. Its quite messy and ive spent ages looking for a way to do it. The Q was differentiate with respect to x and simplify where possible

$sin^3xcos3x+cos^3xsin3x$

I got down to: $\frac{sin^3xsin3x}{3}+3sin^2xcos3xcosx+\frac{cos^3 xcos3x}{3}-sin3xcos^2xsinx$ a bit long dont you think?

It's a bit completely incorrect is what I really think! (and perhaps not long enough.) You seem to be confusing differentiation with anti-differentiation. The derivative of cos 3x, for example is -sin(3x) times 3, not divided by 3.

The derivative of [tex]sin^3(x) cos(3x)[/itex] alone is $(sin^3(x))' cos(3x)+ sin^3(x)(cos(3x))'= (3 sin^2(x)cos(x))(cos(3x))- 3sin^3(x)sin(3x)$ and you do the derivative of $cos^3(x) sin(3x)$ the same way.

**mr fantastic** · November 6th 2008, 05:19 AM

Originally Posted by oxrigby

Im doing this differential and used the chain/product rule up untuil i cannot simplify any further. Its quite messy and ive spent ages looking for a way to do it. The Q was differentiate with respect to x and simplify where possible

$sin^3xcos3x+cos^3xsin3x$

I got down to: $\frac{sin^3xsin3x}{3}+3sin^2xcos3xcosx+\frac{cos^3 xcos3x}{3}-sin3xcos^2xsinx$ a bit long dont you think?

There are a number of careless mistakes (corrected in red below) in your answer.

${\color{red}-3 }sin^3x sin3x+3sin^2xcos3xcosx+ {\color{red}3} cos^3xcos3x - {\color{red}3} sin3xcos^2xsinx$

It's still long, which is why simplification is asked for.

**Soroban** · November 6th 2008, 05:29 AM

Hello, oxrigby!

I have no idea how you're getting fractions . . .

Differentiate: . $f(x) \:=\:\sin^3\!x\cdot\cos3x + \cos^3\!x\cdot\sin3x$

Product Rule and Chain Rule . . .

$f'(x) \;=\;(\sin^3\!x)\!\cdot\!(\text{-}3\sin3x) + (3\sin^2\!x\cos x)\!\cdot\!(\cos3x) +$ $(\cos^3\!x)\!\cdot\!(3\cos3x) + (\text{-}3\cos^2\!3x\sin x)\!\cdot\!(\sin3x)$

. . . $= \;-3\sin^3\!x\sin3x + 3\sin^2\!x\cos x\cos3x + 3\cos^3\!x\cos3x - 3\sin x\cos^2\!x\sin3x$

$\text{Factor: }\;3\sin^2\!x\cdot\underbrace{(-\sin x\sin3x \;+\; \cos x\cos3x)} \;+\; 3\cos^2\!x\cdot\underbrace{(\cos x\cos3x \;-\; \sin x\sin3x)}$
. . . . . . . . . . . . . . . . . . . . . . . . $\nwarrow\;_{\text{ common factor! }}\;\nearrow$

$\text{Factor: }\;(3\sin^2\!x + 3\cos^2\!x)\,(\cos x\cos3x - \sin x\sin3x)$

$\text{Factor: }\;3\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}\,\underbrace{(\cos x\cos3x - \sin x\sin3x)}_{\text{This is }\cos(x + 3x)}$

Therefore: . $f'(x) \;=\;3\cos4x$

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