Derivatives. Product rule with 3 products

EvilBunny
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Homework Statement



If f(x) = (3 x )(sin x) (cos x), find f'( x ).


A question I have is , is there anything special to do when you have 3 products instead of 2



The Attempt at a Solution



Well I used the product rule as if am multipling

(3xsinx) (cosx)

but that doesn't seem to get me the answer or maybe Its something about my answer because I put answers in a computer so sometimes its the notation.

here is my final answer
3sin(x)+cos(x)*3x*cos(x)-sin(x)(3x)sin(x)
 
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well for 3 products...just take take the product of 2 terms and multiply by the differential of the of the 3rd term...if you don't get it

\frac{d}{dx}(UVW)=UV\frac{dW}{dx}+UW\frac{dV}{dx}+VW\frac{dU}{dx}
 
K managed to get it off of that thx
 
although the equation above is impressive and simple. I hate to learn yet another differentiation rule. your initial approach is correct.

(3xsinx) (cosx) = [(3x sinx) (-sin x)]+ [first ' * (cosx)]

the same old (first * second ') + (first ' * second) product rule.

now take derivative of (3x * sin x) with the product rule and plug it in where first' goes.

your computer probably does a better job of simplification than u :)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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