Trig Derivative Problem.

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February 4th 2013, 08:48 PM
KhanDisciple

Trig Derivative Problem.

Hello everyone, any assistance would be greatly appreciated.

Let $f(x)$ $=$ $12xsinxcosx$ . Find $f'(\frac{-3\pi}{2})$ .

I attempt to differentiate the function using the product rule:
I use the product rule for $sinx cosx$
$sinx(-sinx)+cosx(cosx)=-sin^2x+cos^2x$

And then $12x(-sin^2x+cos^2x)$
$=12x(-1)+(-sin^2x+cos^2x)(12)$
$=-12x-12sin^2x+12cos^2x$

Using the derivative of $f(x)$ , I try to evaluate $f'(\frac{-3\pi}{2})$ .
$-12(\frac{-3\pi}{2})-12sin^2(\frac{-3\pi}{2})+12cos^2(\frac{-3\pi}{2})$

I end up getting $18\pi-12$

However, the book answer is just $18\pi$ .
Can someone tell me where I went wrong? Thanks in advance.
February 4th 2013, 09:24 PM
Prove It

Re: Trig Derivative Problem.

You're best off if you use the identity $\displaystyle \begin{align*} \sin{(2x)} \equiv 2\sin{(x)}\cos{(x)} \end{align*}$ . Then $\displaystyle \begin{align*} f(x) = 12x\sin{(x)}\cos{(x)} = 6x\sin{(2x)} \end{align*}$ . Then by the product rule we have

$\displaystyle \begin{align*} f(x) &= 6x\sin{(2x)} \\ f'(x) &= 6\left[ 2x\cos{(2x)} + \sin{(2x)} \right] \\ f'\left( -\frac{3\pi}{2} \right) &= 6\left[ 2\left( -\frac{3\pi}{2} \right) \cos{( -3\pi )} + \sin{(-3\pi)} \right] \\ &= 6\left[ -3\pi \left( -1 \right) + 0 \right] \\ &= 18\pi \end{align*}$
February 5th 2013, 08:22 AM
KhanDisciple

Re: Trig Derivative Problem.

Thank you very much for the quick response Prove It.

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