Derivatives. Product rule with 3 products

EvilBunny
Messages
39
Reaction score
0

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)
 
Physics news on Phys.org
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 :)
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top